A 15-meter by 23-meter garden is divided into two sections. Two sidewalks run along the diagonal of the square section and along the diagonal of the smaller rectangular section.

A rectangle is shown. The length of the top and bottom sides is 23 meters. The length of the left and right sides is 15 meters. A vertical line is drawn from the top to bottom side to split the shape into a square and a rectangle. Lines are drawn from the bottom left point to the top of the vertical line and from the bottom right point to the top of the vertical line.

What is the approximate sum of the lengths of the two sidewalks, shown as dotted lines?

21.2 m
27.5 m
32.5 m
38.2 m

The total amount spent by Ivan is given by the equation A = $ 40.20

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the total amount spent by Ivan be represented as A

Now , the equation will be

The cost of one large necklace = $ 5.90

The cost of one small necklace = $ 4.90

The number of large necklaces = 1

The number of small necklaces = 7

So , the cost of 7 small necklaces = 7 x cost of one small necklace

Substituting the values in the equation , we get

The cost of 7 small necklaces = 7 x $ 4.90

The cost of 7 small necklaces = $ 34.30

And , the total amount spent by Ivan = cost of one large necklace + cost of 7 small necklaces

On simplifying the equation , we get

The total amount spent by Ivan A = 5.90 + 34.30

The total amount spent by Ivan A = $ 40.20

Hence , the equation is A = $ 40.20

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Answer:

Step-by-step explanation:

ito answer: Combine 130 and x2.C=x230−2x+1530

Answer:

-0.4444444444 or just -0.4

30. The pair of angles equal in measurement are ∠1 and ∠9.

31. The sum of the angles x and y is 165 degrees.

When parallel lines are cut by a transversal line, angle relationships are formed such as corresponding angles, alternate interior angles, alternate exterior angles, linear angles, vertically opposite angles etc.

Therefore, let's use the angle relationships to find the missing angles in the parallel lines.

Therefore, ∠1 and ∠9 are corresponding angles. Hence, they are congruent.

Let's find the angles x and y to know the sum of the angles.

Therefore,

x = 55 degrees(corresponding angles)

y = 180 - 70 = 110(sum of angles on a straight line)

Therefore,

x + y  = 55 + 110 = 165 degrees.

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As two distinct ordered pairs (6, 5) and (6, 10) have the same first coordinate the relation is not a function.

A function can be defined as the outputs for a given set of inputs.

The inputs of a function are known as the independent variable and the outputs of a function are known as the dependent variable.

We know a relation is a function if it is one-to-one and many-to-one.

And, A relation is not a function if it is one to many.

From the given arrow diagram.

The domain of the relation is, {3, 6, 12, 18} and the range of the relation is

{5, 7, 9, 10}.

But preimage 6 has two different images namely 5 and 10 so the relation is not a function.

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Answer:

18.2

Step-by-step explanation:

replace where x is with 3

[tex]3 + \frac{1}{2} + (5 \times 3) - \frac{3}{10} [/tex]

[tex]3 \frac{1}{2} + 15 - \frac{3}{10} [/tex]

[tex]15 - \frac{3}{10} = 14 \frac{7}{10} [/tex]

[tex]3 \frac{1}{2} + 14 \frac{7}{10} = 18 \frac{1}{5} [/tex]

[tex]3.5 + 15 - 0.3 = 18.2[/tex]

Answer: 5 (v+1)

Step-by-step explanation:

( i_i )

$1800 is the amount invested

An interest rate is the amount of interest due per period, as a proportion of the amount lent, deposited, or borrowed. The total interest on an amount lent or borrowed depends on the principal sum, the interest rate, the compounding frequency, and the length of time over which it is lent, deposited, or borrowed.

Nico invests some money at 8%;

let M = amount invested

Nico also invests M - 800 at 7%

After 1 year, the total interest is $214.

M (0.08) + (M - 800)0.07 = $214

0.08M + 0.07M - 56 = $214

0.15M - 56 =  $214

0.15M = $160 + $56

0.15M = $270

Divide both sides of the equation by 0.15

M = $1800

Hence, the amount invested is $1800

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Answer:

1.a)  x = g(2)

1.b)  See below

2.a)  See below

2.b) α = 60°

3.a)  2 + h

3.b)  2

4.a)  Interval [3, 4] = 3 m/s

       Interval [3, 3.1] = 2.1 m/s

4.b)  2 m/s

Step-by-step explanation:

Part (a)

Given function f(x):

[tex]f(x)=\dfrac{x}{x-2}[/tex]

If the range of f is (-∞, 1) U (1, ∞), then a value of y in the range of f is y=2.

Set the function equal to the value of y and solve for x:

[tex]\begin{aligned}y=2 \implies \dfrac{x}{x-2}&=2\\x&=2(x-2)\\x&=2x-4\\-x&=-4\\x&=4\end{aligned}[/tex]

The solution written as a function x = g(y) is:

Part (b)

Given function g(y):

[tex]g(y)=\dfrac{2y}{y-1}[/tex]

Verify that g is the inverse function of f by calculating g(f(x)):

[tex]\begin{aligned}\implies g(f(x)) &=\dfrac{2f(x)}{f(x)-1}\\\\&=\dfrac{2\left(\frac{x}{x-2}\right)}{\left(\frac{x}{x-2}\right)-1}\\\\&=\dfrac{2\left(\frac{x}{x-2}\right)}{\left(\frac{x-(x-2)}{x-2}\right)}\\\\&=\dfrac{\left(\dfrac{2x}{x-2}\right)}{\left(\dfrac{2}{x-2}\right)}\\\\&=\dfrac{2x}{2}\\\\&=x \quad (\text{for all $x\neq 2$})\end{aligned}[/tex]

Similarly, calculate f(g(y)):

[tex]\begin{aligned}\implies f(g(y))&=\dfrac{g(y)}{g(y)-2}\\\\&=\dfrac{\left(\frac{2y}{y-1}\right)}{\left(\frac{2y}{y-1}\right)-2}\\\\&=\dfrac{\left(\frac{2y}{y-1}\right)}{\left(\frac{2y-2(y-1)}{y-1}\right)}\\\\&=\dfrac{\left(\dfrac{2y}{y-1}\right)}{\left(\dfrac{2}{y-1}\right)}\\\\&=\dfrac{2y}{2}\\\\&=y \quad \text{(for all $y \neq 1$)}\end{aligned}[/tex]

Hence verifying that g is the inverse function of f.

[tex]\boxed{\begin{minipage}{5 cm}\underline{Trigonometric Identities}\\\\$\tan x=\dfrac{\sin x}{\cos x}$\\\\\\$\cos \left(\arctan \left(x\right)\right)=\dfrac{1}{\sqrt{1+x^2}}$\\\\\\$\sin\left(\arctan \left(x\right)\right)=\dfrac{x}{\sqrt{1+x^2}}$\\\end{minipage}}[/tex]

Part (a)

If  y = arctan(x)  then  tan(y) = x.

Use the trigonometric identities to express sin(y) and cos(y) in terms of x.

[tex]\boxed{\begin{aligned}\tan(y) &= x\\\implies \dfrac{\sin y}{\cos y}&=x\\\sin y&=x \cos y\\ \sin y&=x \cos (\arctan x)\\ \sin y&=\dfrac{x}{\sqrt{1+x^2}}\end{aligned}}[/tex]

[tex]\boxed{\begin{aligned}\tan(y) &= x\\\\\implies \dfrac{\sin y}{\cos y}&=x\\\\\dfrac{\sin y}{x}&= \cos y\\\\ \cos y&=\dfrac{\sin(\arctan x)}{x}\\\\ \cos y&=\dfrac{\dfrac{x}{\sqrt{1+x^2}}}{x}\\\\\cos y&=\dfrac{x}{\sqrt{x^2+1}x}\\\\\cos y&=\dfrac{1}{\sqrt{x^2+1}}\end{aligned}}[/tex]

Part (b)

Given linear equation:

[tex]y=\sqrt{3}x+1[/tex]

As the slope of a linear equation y = mx + b is m, then the slope (m) of the given line is √3.

If m = tan(α) then:

[tex]\begin{aligned}\implies m&=\sqrt{3}\\\tan (\alpha)&=\sqrt{3}\\\alpha&=\arctan \sqrt{3}\\\alpha&=60^{\circ}+180^{\circ}n\\\end{aligned}[/tex]

If α is the angle the line forms with the x-axis, then α = 60°.

Given function:

[tex]f(x)=x^2-1[/tex]

Part (a)

[tex]\begin{aligned}\dfrac{f(1+h)-f(1)}{h} &=\dfrac{((1+h)^2-1)-(1^2-1)}{h}\\\\&=\dfrac{(1+2h+h^2-1)-0}{h}\\\\&=\dfrac{2h+h^2}{h}\\\\&=2+h\end{aligned}[/tex]

Part (b)

Part a used differentiating from first principles to find the gradient of f(x) at x = 1.  As h gets smaller, the gradient of the straight line gets closer and closer to the gradient of the curve.  Therefore, as h gets close to zero, (2 + h) gets close to 2.  Therefore, the slope of the tangent line to the graph of f(x) at the point where x = 1 is 2.

A particle is moving on the x-axis and its position at time is given by

[tex]x(t)=t^2-4t+3[/tex]

Evaluate the position of the particle at t = 3, t = 4 and t = 3.1:

[tex]x(3)=3^2-4(3)+3=0[/tex]

[tex]x(4)=4^2-4(4)+3=3[/tex]

[tex]x(3.1)=(3.1)^2-4(3.1)+3=0.21[/tex]

Average velocity formula

[tex]\overline{v}=\dfrac{s(t_2)-s(t_1)}{t_2-t_1}[/tex]

Therefore, the average velocity of this particle over the interval [3, 4]:

[tex]\overline{v}=\dfrac{x(4)-x(3)}{4-3}=\dfrac{3-0}{1}=3\;\; \rm m/s[/tex]

The average velocity over the interval [3, 3.1] is:

[tex]\overline{v}=\dfrac{x(3.1)-x(3)}{3.1-3}=\dfrac{0.21-0}{0.1}=2.1\;\; \rm m/s[/tex]

Part (b)

To find an equation for velocity, differentiate the given equation for displacement:

[tex]v(t)=\dfrac{\text{d}x}{\text{d}t}=2t-4[/tex]

To calculate the instantaneous velocity of the particle at time t = 3, substitute t = 3 into the equation for velocity:

[tex]v(3)=2(3)-4=2\;\; \rm m/s[/tex]

Answer:

26

Step-by-step explanation:

2 squared equals 4. So 30 - 4 = 26

hope it helps!

Solve the equation y = f(x) to determine the values of the independent variable x and obtain the domain.

What is meant by domain and range?

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Solve the equation y = f(x) to determine the values of the independent variable x and obtain the domain.

The domain of a function is the set of values that we are allowed to plug into our function. This set is the x values in a function such as f(x). The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in.

All of the values that can go into a relation or function (input) are called the domain. All of the values that come out of a relation or function (output) are called the range.

To find the domain and range, we simply solve the equation y = f(x) to determine the values of the independent variable x and obtain the domain. To calculate the range of the function, we simply express x as x=g(y) and then find the domain of g(y).

Given equation: y=f(x)

We can solve for the value of x by rearranging the equation to the form x = [tex]f^{-1y}[/tex]

Domain:

x = [tex]f^{-1y}[/tex]

The domain of f(x) is all the possible values of x for which y is defined.

Range:

y = g(x)

The range of g(x) is all the possible values of y for which x is defined.

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Answer:

x=3

Y=−4

Step-by-step explanation:

Y=−7/3x+3

Y=−1/3x−3

​

Consider the first equation. Add  

3

7

​

x to both sides.

Y+

3

7

​

x=3

Consider the second equation. Add  

3

1

​

x to both sides.

Y+

3

1

​

x=−3

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

Y+

3

7

​

x=3,Y+

3

1

​

x=−3

Choose one of the equations and solve it for Y by isolating Y on the left hand side of the equal sign.

Y+

3

7

​

x=3

Subtract  

3

7x

​

 from both sides of the equation.

Y=−

3

7

​

x+3

Substitute −

3

7x

​

+3 for Y in the other equation, Y+

3

1

​

x=−3.

−

3

7

​

x+3+

3

1

​

x=−3

Add −

3

7x

​

 to  

3

x

​

.

−2x+3=−3

Subtract 3 from both sides of the equation.

−2x=−6

Divide both sides by −2.

x=3

Substitute 3 for x in Y=−

3

7

​

x+3. Because the resulting equation contains only one variable, you can solve for Y directly.

Y=−

3

7

​

×3+3

Multiply −

3

7

​

 times 3.

Y=−7+3

Add 3 to −7.

Y=−4

The system is now solved.

Y=−4,x=3

Answer:

The answer is 1720ㅤㅤㅤㅤㅤㅤㅤㅤ

Answer:

Step-by-step explanation:

To find the x-coordinates of the points where the curve with the equation y = x³ - 4x² + 5x + 4 has a gradient of 1, we need to find the points where the derivative of the equation y = x³ - 4x² + 5x + 4 is equal to 1. The derivative of the equation y = x³ - 4x² + 5x + 4 is given by:

dy/dx = 3x² - 8x + 5

So, we need to find the solutions of the equation:

3x² - 8x + 5 = 1

This is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. To use the quadratic formula, we can write:

x = (-b ± √(b² - 4ac)) / (2a)

Where a = 3, b = -8, and c = 4. Plugging in these values, we get:

x = (-(-8) ± √((-8)² - 4 * 3 * 4)) / (2 * 3)

x = (8 ± √(64 - 36)) / 6

x = (8 ± √(28)) / 6

x = (8 ± 2√7) / 6

So, the two solutions are:

x = (8 + 2√7) / 6 and x = (8 - 2√7) / 6

These are the x-coordinates of the points where the curve with the equation y = x³ - 4x² + 5x + 4 has a gradient of 1.

Answer:

x = -1/18

Step-by-step explanation:

[tex]\frac{5}{6} +6x=\frac{1}{2}[/tex]

[tex]6x=\frac{1}{2} -\frac{5}{6}[/tex]

[tex]6x=\frac{6-10}{12}[/tex]

[tex]6x=\frac{-4}{12}[/tex]

[tex]6x=-\frac{1}{3}[/tex]

[tex]x=\frac{-\frac{1}{3} }{6} =\frac{-1}{(3)(6)}[/tex]

[tex]x=-\frac{1}{18}[/tex]

Hope this helps

The maximum possible cost of a portion of fish is £4.

The cost of a product or service is the amount of money spent on it. It is the whole cost of purchasing, manufacturing, or producing something. The cost of a work may also refer to the amount of time and effort required to finish it. When determining whether or not to buy a product or service, cost is a crucial consideration, and it may be influenced by a number of variables such as material prices, labor expenses, overhead, and demand.

A piece of fish can cost no more than £4.

We may use the following equation to figure this out:

5 servings of fish at £f each + 2 bags of chips at £2 each = £20

This equation may then be rearranged to solve for f:

f = (20 - (2 x 2))/5

f = 16/5

f = £3.20

As a result, the most a slice of fish may cost is £4.

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A salesman earns $490 as commission on his sales.

Percentage is defined as a given part or amount in every hundred. It is a fraction with 100 as the denominator and is represented by the symbol "%".

Given that, a salesman earns 7% commission on all the merchandise that he sells.

Last month he sold $7000 worth of merchandise.

Now, the commission is 7% of 7000

7/100 ×7000

= 0.07×7000

= $490

Therefore, a salesman earns $490.

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The Central Limit Theorem states that if the population is normally distributed and the sample size is reasonably large, then the sample mean will be normally distributed.

The Central Limit Theorem states that if the sample size is reasonably large and the population is normally distributed, then the sample mean will be normally distributed. This means that we can use the normal distribution to make inferences about the population mean. Specifically, it allows us to calculate the probability that the population mean falls within a certain range, given the sample mean. This is useful because we can use the sample data to make assumptions about the population as a whole. For example, if the sample data is normally distributed, we can assume that the population mean is also normally distributed. This is especially useful when the population size is large and it would be difficult to collect data from every individual in the population. The Central Limit Theorem gives us an efficient way to make inferences about the population mean.

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Answer: Find the length of one side and square it

Example: Lets say one side of the square has a length of 7. The area would be 49 because 7x7=49

The exponential expression in the format cx^n is given as follows:

2.5x^(-6).

The expression for this problem is defined as follows:

5/(2x^6).

The desired format is given as follows:

cx^n.

For the parameter c, we must simply divide the numeric coefficients of the numerator and the denominator, hence:

5/2 = 2.5.

For the parameter n, we have the positive exponent in the denominator, hence it can be written as a negative exponent in the numerator, that is:

n = -6.

Hence the expression is given as follows:

2.5n^(-6).

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Step-by-step explanation:

Factor the denominator to simplify the expression: x^2 + x - 2 = (x - 1)(x + 2)

Use partial fraction decomposition to write the integrand as the sum of simpler terms: (x^3 + 2x^2 + 3x + 4)/(x^2 + x - 2) = A/(x - 1) + B/(x + 2) + Cx + D, where A, B, C, and D are constants.

Multiply both sides of the equation by (x^2 + x - 2) to find the values of A, B, C, and D: x^3 + 2x^2 + 3x + 4 = A(x + 2) + B(x - 1) + (Cx + D)(x^2 + x - 2)

Substitute x = 1 and x = -2 into the equation to find two equations for A, B, C, and D:

x = 1: 4 + 2 + 3 + 4 = 9 = A(-2) + B + (C + D)(-1)

x = -2: -8 - 4 + 6 - 8 = -16 = A(1) + B(-2) + (C - 2D)(4)

Solve the system of equations to find the values of A, B, C, and D:

A = 9/3, B = 5, C = -11/3, D = 7/3

Use the partial fraction decomposition to integrate the integrand:

Integral of 1/(x - 1) = ln|x - 1|

Integral of 1/(x + 2) = ln|x + 2|

Integral of x = x^2/2

Integral of 1 = x

Evaluate the definite integral by subtracting the values of the antiderivatives at the limits of integration:

Integral from x=0 to x=1 of (x^3 + 2x^2 + 3x + 4)/(x^2 + x - 2) = (ln|x + 2| - ln|x - 1| + x^2/2 - x)|x=1 - x=0 = (ln(3) - ln(-1) + 1/2 - 1) = (ln(3) + ln(1) + 1/2) = ln(3) + 1.

So the definite integral of (x^3 + 2x^2 + 3x + 4)/(x^2 + x - 2) from x=0 to x=1 is equal to ln(3) + 1.

Answer:

Step-by-step explanation:

The height of the right cone will be 57 units.

The area a cone takes up in a three-dimensional plane is known as its volume. A cone's base is circular, so it is made of a radius and a diameter.

Given that the volume of a right cone is 4275π cubic units. The circumference of the cone is 30π.

The height of the cone will be calculated as below:-

First, calculate the radius of the base of the cone,

Circumference = 2πr

30π = 2πr

r = 15 units

The height of the cone:-

Volume = 4275π

( 1 / 3 ) πr²h =  4275π

h = ( 4275 x 3 ) / r²

h = ( 4275 x 3 ) / ( 15)²

h = 57 units

Hence, the height will be 57 units.

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The amount after 15 years with interest compounded quarterly is given by the equation A = $ 4,231.79

Compound interest is interest based on the initial principle plus all prior periods' accumulated interest. The power of compound interest is the ability to generate "interest on interest." Interest can be added at any time, from continuously to daily to annually.

The formula for calculating Compound Interest is

A = P ( 1 + r/n )ⁿᵇ

where A = Final Amount

P = Principal

r = rate of interest

n = number of times interest is applied

b = number of time periods elapsed

Given data ,

Let the amount after 15 years be represented as A

Now , the equation will be

Let the principal amount be P = $ 3000

Let the number of years be b = 15 years

Let the number of times interest is applied be n = 4

Let the percentage of interest be r = 2.3 %

And , A = P ( 1 + r/n )ⁿᵇ

Substituting the values in the equation , we get

A = 3000 ( 1 + 0.023/4 )⁴ˣ¹⁵

A = 3000 ( 1 + 0.00575 )⁶⁰

On further simplification , we get

A = 3000 ( 1.00575 )⁶⁰

A = $ 4,231.79

Therefore , the compound interest is I = $ 1,231.49

Hence , the amount after 15 years is $ 4,231.79

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The differential equation (y^2 cos(x−y) −x^2 sin(x−y)) dx+(2xycos(x−y) +x^2 sin(x−y)) dy =0 is not exact, because it cannot be written in the form F(x,y)dx + G(x,y)dy = 0 for some functions F and G.

b. The differential equation (2x+12y^2)dx+(12x+2y)dy=0 is exact, because it can be written in the form F(x,y)dx + G(x,y)dy = 0 for F(x,y) = 2x + 12y^2 and G(x,y) = 12x + 2y.

To find a general solution in an implicit form, we can integrate both sides:

∫ (2x + 12y^2) dx + ∫ (12x + 2y)dy = C

x^2 + 6y^2 + 12xy = C

c. The differential equation (12x^2 + 2y) dx+(2x + 12y^2)dy=0 is not exact, because it cannot be written in the form F(x,y)dx + G(x,y)dy = 0 for some functions F and G.

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Using the general formula for the mean, we can see that the mean between 0.6149 and 0.6236 is 0.61925

Suppose we have any set of N numbers:

{x₁, x₂, ..., xₙ}

The mean of these values is given by:

mean = (x₁ + x₂ + ... + xₙ)/N

In this case we want to find the mean between two numbers,  0.6149 to 0.6236

This will be:

mean = (0.6149 + 0.6236)/2 = 0.61925

We can see that the mean between  0.6149 and 0.6236 is 0.61925

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Answer:

Given Figure has set of Stacked cubes.

To Draw the Base Plan, lets see its right side, Left Side and Front side view.

Right side:

It has 3 cubes in base.

2 cubes in 2nd layer clearly seen in figure and only 1 cube in 3rd layer.

Left Side:

It has 2 Cubes in base.

Both Cubes attached to 1st and 2nd cube of right side cubes.

Front Side:

It can be seen that, here base only has 2 rows of cubes.

Therefore,  1st Figure is correct Base Plan (or enclosed figure)

Note: Numbers in enclosed figure just shows the total no. of cubes in that stack.

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The two equations which can be used to represent the situation are;

r + s = 1787

4r + 3s = 5,792

The correct answer choice is option B and E

Reserved seat for basketball game = r

Students seat for basketball game = s

Cost of reserved seat tickets = $4

Cost of students tickets = $3

Total number of people who attended the basketball game= 1,787 people

Total amount earned for tickets sales= $5,792

r + s = 1787

4r + 3s = 5,792

Therefore, the basketball game situation can be represented by the equation r + s = 1787; 4r + 3s = 5,792

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Answer:

y=4(x−2) ^2−14

Step-by-step explanation:

The difference between density-dependent and independent is given below.

By controlling population density-related factors including disease, competition, and prediction, density dependence regulates the population.

A huge population is often where density-dependent operates.

The rate of increase and decrease determines this.

Food, housing, prediction, competition, and disease are factors that depend on population density.

Density Independence: Natural disasters and the weather are examples of things that govern the population without taking density into account.

It is possible to use density independence in both small and big populations.

Both small and large populations can be served by density-independent.

Fluctuations, fires, droughts, extremely high temperatures, and tornadoes are examples of density-independent variables.

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