after finishing the prep work, gilberta and maría start removing wallpaper at the same time. gilberta removes the paper at a constant rate of 4.3\,\text{m}^24.3m 2 4, point, 3, start text, m, end text, squared per hour, while maría removes 3.4\,\text{m}^23.4m 2 3, point, 4, start text, m, end text, squared of paper per hour. gilberta's room starts with 35\,\text{m}^235m 2 35, start text, m, end text, squared of paper, and maría's room starts with 30.5\,\text{m}^230.5m 2 30, point, 5, start text, m, end text, squared of paper.

Answer:

M&M's for =Rs.8 per bag

Both team sells a total 240 of a total of Rs.1614

Now

The total of both =1614-240=1374

The basketball team sold 102 packages of licorice and 138 bags of M&M's.

Given that the packages of licorice sold for $5 per package and that of the bags of M&M's for $8 per bag.

The total selling was total of 240 of both for a total of $1614.

We need to find the number of packages of licorice and bags of M&M's were sold.

Let's assume that the number of packages of licorice sold is represented by the variable "L" and the number of bags of M&M's sold is represented by the variable "M."

From the given information, we have two pieces of information to form equations:

The total number of packages and bags sold: L + M = 240

The total revenue from selling these items: 5L + 8M = 1614

Now, we can solve these two equations simultaneously to find the values of L and M.

First, let's solve the first equation for one of the variables (let's solve for L):

L = 240 - M

Now, substitute this expression for L in the second equation:

5(240 - M) + 8M = 1614

Now, distribute the 5:

1200 - 5M + 8M = 1614

Combine the M terms:

3M = 1614 - 1200

3M = 414

Now, solve for M:

M = 414 / 3

M = 138

Now that we know the value of M, we can find the value of L using the first equation:

L = 240 - M

L = 240 - 138

L = 102

So, the basketball team sold 102 packages of licorice and 138 bags of M&M's.

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The equation system to represent these two plans is:

c=100+10d, c=40d where c is the total cost and d represents the data used.

We have to understand the plans provided by AT&K and Verikon separately to find the equation system. We know that the total cost incurred is c and the amount of data used is d. It is given that AT&K charges a monthly flat rate of $100 and additional $10 for every gigabyte of data used. We can say that for a month the total cost incurred from AT&K is $100 plus $10xd, the amount of data used.

c=100+10d

For verikon it is given that it does not have any flat monthly rate and it charges $40 for each gigabyte of data used.We can easily say that the total cost in this case is $40 multiplied to amount of data used.

c=40d

Thus the system of equation for the two different plans is

c=100+10d and c=40d

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Small, random genetic changes that occur over generations is known as genetic drift.

Genetic drift is an evolutionary process in which the frequency of a given gene allele varies randomly across a population. The impacts of genetic drift can be severe, often causing traits to become excessively common or to vanish from a community, even though it predominantly affects small, isolated populations.

Some people might merely by coincidence leave behind a few more offspring (and genes, of course!) than others in each generation. The next generation's DNA and other genetic components will come from the "fortunate" people, not necessarily the healthier or "better" people.

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[tex]- 4 - 9 = - 13[/tex]

Answer:

no

Step-by-step explanation:

x=5

5(5+2)=35

8(5)-3=37

37 does not equal 35

[tex]Answer: Q\ is\ not\ midpoint\ of\ \overline {PR}[/tex]

Step-by-step explanation:

[tex]\displaystyle\\\overline {PQ}+\overline {QR}=\overline {PR}\\5(x+2)+(8x-3)=72[/tex]

Expand the parentheses on the left side of the equation:

x(5)+2*(5)+8x-3=72

5x+10+8x-3=72

13x+7=72

13x+7-7=72-7

13x=65

Divide both parts of the equation by 13:

x=5

Hence,

5(5+2)=

5(5)+2(5)=

25+10=

35

8(5)-3=

40-3=

37

35≠37

Answer:

4725

Step-by-step explanation:

105 students

think of 100 students

100 × 45 = 4500

now the remaining 5 students

cost for 10 students will be

45 × 10 = 450

divide 450 by 2 to find out for 5 students

it is 225

4500 + 225 = 4725

Weight of pet & age of pet  are   the two categories that would yield quantitative data.

What foundation does qualitative data have?

Which two sorts of quantitative data are there?

Weight of pet & age of pet  are   the two categories that would yield quantitative data because both quantity can measured or counted.

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Answer:

See attached graph for lines

Left side line is y = -3x-2

Right side line is y = 2x + 8

Point of intersection is at (-2,4)

Step-by-step explanation:

The line graphs are attached

The solution for (x, y) is the point of intersection  of the two lines

This is at point (2,4)

We can also solve this algebraically by solving first for x and then y

–3x – 2 = 2x + 8

The two line equations are

y = -3x - 2 and y = 2x+8

3x – 2 = 2x + 8

==> 3x -2x - 2 = 2x-2x + 8   (subtract 2x on both sides)

==>  -5x -2 = 8

==> -5x -2 + 2 = 8 + 2    (add 2 on both sides to isolate the x term)

==> -5x = 10==> -5x/5 = 10/5 (divide by 5 both sides)

==> x = -2

Plug this value of x into the RHS giving

y = 2(-2) + 8 ==> y = -4 + 8 ==> y = 4

So (-2,4) is the point of intersection of the two lines and is consistent with the point of intersection on the graph

Answer:

Step-by-step explanation: Well €550 Euros is equal to $547.42 US dollars in currency. But since this is approximately, (D) should be correct.

The function rules for each of the given tables are;

11) f(x) = 4x

12) f(x) = x - 3

13) f(x) = x + 12

14) f(x) = x/10

15) This doesn't have a function rule.

16) Function rule here is non existent

11) In this table, we can observe that the output values are all four times the input value. Thus, the function rule here is;

f(x) = 4x

12) In this table, we can observe that the output values are all 3 units less than the input value. Thus, the function rule here is;

f(x) = x - 3

13) In this table, we can observe that the output values are all 12 units more than the input value. Thus, the function rule here is;

f(x) = x + 12

14) In this table, we can observe that the output values are equal to the input value divided by 10. Thus, the function rule here is;

f(x) = x/10

15) This doesn't have a function rule as the values don't match a pattern

16) Function rule here is non existent

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Answer:

x<44

Step-by-step explanation:

Step-by-step explanation:

seven-hundred thirty thousand eight-hundred twelve

Answer:

Hey!

Since we know that both expressions lay on a straight line, we know that when both lines add up it should be 180°

We can write the expression as (7x - 1) + (6x - 1) = 180

Simplify

→ (7x + −1) + (6x + −1) = 180

Combine Like Terms

→ (7x + 6x) + (-1 + -1) = 180 → 13 + -2 = 180

Add 2 To Both Sides

→ 13x - 2 + 2 = 180 + 2 → 13x = 182

Divide 13 To Both Sides

13x/13 = 182/13 → x = 14

Hence, the value of x is 14!

-------------------------------------------------

Hope This Helped! Good Luck!

Answer:

[tex]{ \tt{f(n) = 7n - 5}} \\ { \tt{f( - 2) = 7( - 2) - 5}} \\ { \tt{f( - 2) = - 14 - 5}} \\ { \tt{f( - 2) = - 19}}[/tex]

The inequality which represents all the possible values for x is (2x/3) - 6 ≥ 24.

It is given in the question that the difference of two-thirds of a number x and 6 is at least -24.

We have to find the inequality which represents all the possible values for x.

In mathematics, Inequality is a statement of an order relationship—greater than, greater than or equal to, less than, or less than or equal to—between two numbers or algebraic expressions.

Hence, according to the data given in the question, we can write,

(2x/3) - 6 ≥ 24

Here, we can also find the possible values of x.

(2x/3) - 6 ≥ 24

2x/3  ≥ 24 + 6

2x/3  ≥ 30

x  ≥ 30*3/2

x  ≥ 45

Hence, every number greater than or equal to 45 will be a possible value of x.

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The perimeter of the rectangle with length 8x - 1 and width 10x + 3 is 20x + 6

What is perimeter of the rectangle?

The perimeter of the rectangle with length 8x - 1 and width 10x + 3 is 20x+ 6

The formula for calculating the perimeter of a rectangle is expressed as:

P = 2( l + w)

Given the following parameters

length l = 8x - 1

width w = 2x + 4

Substitute the expressions for the length and width into the formula;

P = 2(8x - 1 + 2x + 4)

P = 2(10x + 3)

Distribute 2 over each term in the parenthesis

P = 2(10x) + 2(3)

P = 20x + 6

Hence the perimeter of the rectangle with length 8x - 1 and width 10x + 3 is 20x + 6.

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The complete question is -

The perimeter formula for a rectangle is 2l + 2w, where l is the length and w is the width. Complete the steps to find an expression that represents the perimeter of the rectangle. Substitute the expressions for length and width into the formula 2l + 2w. Distribute 2 to each term in the parentheses. Combine like terms. inches

Answer:

Step-by-step explanation:

1035x-992/840

The general solution of the given equation is y² = 8x²Inx + cx².

A first-order linear differential equation is one that has the formula:

[tex]\frac{d y}{d x}+p(x) y=q(x)[/tex]

where p(x) and q(x) are continuous functions of x.

We begin by determining the integrating factor for the first-order linear differential equation.

I.F. = [tex]e^{\int p(x) d x}[/tex]

The solution to the first order linear differential equation is:

[tex]y \cdot e^{\int p(x) d x}=\int q(x) \cdot e^{\int p(x) d x} d x+c .[/tex]

The solution to the first order linear differential equation is:

[tex]\int \frac{1}{x} d x=\ln x+c[/tex]

So, given: [tex]y^{\prime}=\frac{\left(4 x^2+y^2\right)}{(x y)}[/tex]

The above differential equation can be written as:

[tex]\begin{aligned}&\frac{d y}{d x}=4 \frac{x}{y}+\frac{y}{x} \\&\frac{d y}{d x}-\frac{y}{x}=\frac{4 x}{y} \\&y \frac{d y}{d x}-\frac{y^2}{x}=4 x\end{aligned}[/tex]

Substitute, [tex]y^2=t, 2 y \frac{d y}{d x}=\frac{d t}{d x}[/tex].

[tex]\begin{aligned}&\frac{1}{2} \frac{d t}{d x}-\frac{t}{x}=4 x \\&\frac{d t}{d x}-\frac{2 t}{x}=8 x\end{aligned}[/tex]

Because this is a first-order linear differential equation, we must first determine the integrating factor:

[tex]\text { I. F. }=e^{-2 \int \frac{1}{x} d x}=e^{-2 \ln x}\\\text { I.F. } F=\frac{1}{x^2}[/tex]

As a result, the following is the general solution to the given IVP:

[tex]$$t \cdot(I . F .)=\int(I . F .) \cdot 8 x d x+c$$$t \cdot\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \cdot 8 x d x+c$$t \cdot\left(\frac{1}{x^2}\right)=8 \int \frac{1}{x} d x+c$$t \cdot\left(\frac{1}{x^2}\right)=8 \ln x+c$$t=8 x^2 \ln x+c x^2$[/tex]

By reversing the value of t:

[tex]y^2=8 x^2 \ln x+c x^2[/tex]

We now test the solution by substituting [tex]y^2=8 x^2 \ln x+c x^2[/tex] it in the original differential equation.

[tex]\begin{aligned}&\frac{d}{d x}\left(y^2\right)=\frac{d}{d x}\left(8 x^2 \ln x+c x^2\right) \\&2 y \frac{d y}{d x}=16 x \ln x+8 x+2 c x \\&\frac{d y}{d x}=\frac{16 x \ln x+8 x+2 c x}{2 y} \\&\frac{d y}{d x}=\frac{8 x \ln x+4 x+c x}{y}\end{aligned}[/tex]

We calculated the value of [tex]y^2[/tex] and [tex]\frac{d y}{d x}[/tex] from the original differential equation:

[tex]\begin{aligned}&\frac{8 x \ln x+4 x+c x}{y}=\frac{4 x^2+8 x^2 \ln x+c x^2}{x y} \\&8 x \ln x+4 x+c x=8 x \ln x+4 x+c x .\end{aligned}[/tex]

Therefore, the general solution of the given equation is y² = 8x²Inx + cx².

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The complete question is given below:
For the following ODE, find a general solution, show the steps of derivation and check your answer by substitution.

[tex]y^{\prime}=\frac{\left(4 x^2+y^2\right)}{(x y)}[/tex]

Answer:

a) 512cm³    b) 2 × 2 ×2

Step-by-step explanation:

A CUBE'S LENGTH WIDTH AND HEIGHT ARE EQUAL

a) Volume of a cube = length × width × height

as the length width and height are equal so it will be

8 × 8 × 8 = 512

b) Volume of a cube = length × width × height

as length, width and height are equal lets suppose the value of each of them will be = x

volume = x × x × x

vol = x³

8 = x³

∛8 = x

x = 2

f(x) + k will shift the graph up k units.

For example, the notation f(x) + 5 will shift the curve up by 5

If k is negative, then the curve shifts down that amount

-------------------------

f(x-h) shifts the graph to the right h units

Example: f(x-2) shifts to the right 2 units

Another example: f(x+7) shifts left 7 units. In this case, h = -7

--------------------------

y = -f(x) is a reflection over the x axis

This is because we're multiplying each y coordinate by -1

A point like (1,2) reflects to (1,-2)

An equation having variables on both sides can be solved by bringing all variables to one side.

An equation with all variables on one side can be solved by simply keeping the variables on the left and simplifying the value on the right. For example,

5x = 6 + 2x;

3x = 6;

x = 2.

For an equation with variables on both sides, just bring all variables on one side. After that, the same process like above is to be followed. For example,

5y - 3 - 3y = 3y + 5 - 3y;

5y - 3y - 3y + 3y = 5 + 3;

2y = 8;

y = 4;

Thus, methods for solving both kinds of equations is similar.

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An equation having variables on both sides can be solved by bringing all variables to one side.

An equation with all variables on one side can be solved by simply keeping the variables on the left and simplifying the value on the right. For example,

5x = 6 + 2x;

3x = 6;

x = 2.

For an equation with variables on both sides, just bring all variables on one side. After that, the same process like above is to be followed. For example,

5y - 3 - 3y = 3y + 5 - 3y;

5y - 3y - 3y + 3y = 5 + 3;

2y = 8;

y = 4;

Thus, methods for solving both kinds of equations is similar.

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Answer:

[tex]\textsf{1.} \quad s=\dfrac{8r}{9}[/tex]

[tex]\textsf{2.} \quad s=4[/tex]

[tex]\textsf{1.} \quad x=y(w+z)[/tex]

[tex]\textsf{2.} \quad x=10[/tex]

Step-by-step explanation:

Question 1

1.  Solve for s:

[tex]\implies 2r = \dfrac{5s}{2}-\dfrac{s}{4}[/tex]

[tex]\implies 2r = s\left(\dfrac{5}{2}-\dfrac{1}{4}\right)[/tex]

[tex]\implies 2r = s\left(\dfrac{10}{4}-\dfrac{1}{4}\right)[/tex]

[tex]\implies 2r = s\left(\dfrac{9}{4}\right)[/tex]

[tex]\implies 4(2r) = 9s[/tex]

[tex]\implies 8r = 9s[/tex]

[tex]\implies s=\dfrac{8r}{9}[/tex]

2.  When r = ⁹/₂ :

[tex]\implies s=\dfrac{8\left(\frac{9}{2}\right)}{9}[/tex]

[tex]\implies \dfrac{8}{9} \times \dfrac{9}{2}[/tex]

[tex]\implies \dfrac{8 \times9}{9 \times2}[/tex]

[tex]\implies s=\dfrac{72}{18}[/tex]

[tex]\implies s=4[/tex]

Question 2

1.  Solve for x:

[tex]\implies w=\dfrac{x}{y}-z[/tex]

[tex]\implies w+z=\dfrac{x}{y}[/tex]

[tex]\implies y(w+z)=x[/tex]

[tex]\implies x=y(w+z)[/tex]

2.  When x = -8,  y = -5,  z = 6:

[tex]\implies x=-5(6+(-8))[/tex]

[tex]\implies x=-5(6-8)[/tex]

[tex]\implies x=-5(-2)[/tex]

[tex]\implies x=10[/tex]