jack graphs how far he plans ton bike over a 3-day charity ride find the slope of the line

Answer:

Adjacent sides.

Step-by-step explanation:

Answer:

To construct a rectangle, you must have the measurement of one side and it's adjacent side.

Step-by-step explanation:

to construct a rectangle you must have one side and :

(1) it’s opposite sides

(2) it’s adjacent sides

(3) two angles

(4) four angles

Answer:

[tex]\dfrac{7}{8} \textsf{ and } -\dfrac{5}{8}[/tex]

Step-by-step explanation:

To find numbers that are [tex]\dfrac{3}{4}[/tex] units distance from [tex]\displaystyle \dfrac{1}{8}[/tex] , add and subtract [tex]\dfrac{3}{4}[/tex] to and from [tex]\displaystyle \dfrac{1}{8}[/tex]

One of the points is at [tex]\displaystyle \dfrac{1}{8} + \dfrac{3}{4}[/tex] units away and the other at [tex]\displaystyle \dfrac{1}{8} - \dfrac{3}{4}[/tex] units away

Answer:

You have the right form (vertex form). All that you have left to do is learn what parts of the equation signify what.

Step-by-step explanation:

Explanation:

In a quadratic function of form y = a

(

x

−

p

)

2

+ q, the axis of symmetry is found at x = p. As a result, in y = -4

(

x

−

8

)

2

+ 3, the axis of symmetry is at x = 8, since the formula is x - p = x - (+8).

The vertex of a parabola is at (p, q). As for the axis of symmetry, the p value is 8, and in this function the q value is 3. So the vertex is at (8,3).

With this information, the x and y intercepts, and a few other points found by plugging in x values and solving for y, you will be able to graph this parabola. Your graph, if done properly, will look like the following:

graph{y = -4(x - 8)^2 + 3 [-20, 20, -10, 10]}

Answer:

OD

Step-by-step explanation:

move the -8 to the right, and change its sign

x > 5

Answer:

Movie duration:1.5hrs

Answer:

x < 3

Step-by-step explain: I used quizlet

C.78893

Step-by-step explanation:

-_-_+_(*;#-✓✓

Answer:

the slope of the table is -0.857

Answer: -10s +15

-10 = 10 meters down

s = seconds

+ 15 = 15 meters risen

On 4/5 gallon of gasoline, hybrid car will travel a distance of 48 miles.

Suppose, non hybrid car travel "x" distance in 1 gallon of gasoline

Then, hybrid will travel " 1 ⁹/₁₁ * x" distance in 1 gallon of gasoline

Given that, non hybrid car travel distance per gallon is; x = 33 miles per gallon

Then, hybrid car travel,  1 ⁹/₁₁ * x =  1 ⁹/₁₁ * x * 33 = 60 miles per gallon

On 1 gallon of gasoline hybrid car travel 60 miles

On 4/5 gallon of gasoline hybrid car will travel = 4/5 * 60 = 48 miles

On 4/5 gallon of gasoline, hybrid car will travel 48 miles of distance.

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The greatest common factor of 18 12 16 is the number 2.

Answer:

(9c) - 4=5

Step-by-step explanation:

Product means 'The answer when you multiply' So we are working in multiplication. ( x )

9c is the multiplication.

It says 'less' which means subtracting to our answer. So we subtract 4. So far our equation is: (9c) - 4 (Brackets are needed as we are multiplying 9 by c and not multiplying 9 by c - 4)

Lastly, it says the answer must be 5 so we add that at the end. So the answer is:

(9c) - 4=5

Using proportions, it is found that the exact value of 41% of 379 is of 155.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.

The proportion of 41% of 379 is given as follows:

0.41 x 379 = 155.39.

Since 0.39 < 0.5, the exact value is rounded to 155.

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Answer:

[tex]xy\left(7.2x-0.9y\right)[/tex]

Step-by-step explanation:

[tex]Rule: a^{b+c}=a^ba^c\\(x^2y=xxy,\:xy^2=xyy)\\=7.2x(xy)-0.9xyy\\=xy\left(7.2x-0.9y\right)[/tex]

Answer:

the answer is two because ur multiplying hope this helps

Answer:

A because of the way the pipe is designed

Step-by-step explanation:

with ABCD and ABDE being parallelograms, this means a indicated by the marks

AB = ED = CD

and with BD = DF that means that the line segments BF and CE are intersecting each other at their midpoint.

so they represent a pair of diagonals of a parallelogram, which makes BCFE a parallelogram.

Answer: total household present are 136

Step-by-step explanation:

Given:

82.1% of Cincinnati households have a computer.

111.804 households in Cincinnati have a computer.

According to ratio and proportions we have:

let total number of houses be x;

then

(111.804/x)*100=percentage of Cincinnati households having  a computer.    [[tex]gain/total*100 =gain %[/tex]]

then

(111.804/x)*100=82.1

cross multiplying we have

x=(111.804*100)/82.1

x=136 households  (Round to the nearest whole number.)

so total household present are 136

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Answer and Explanation:

The problem provides the following information:

Hypotenuse

=

h

y

p

=

5129

ft

Vertical drop

=

o

p

p

=

1275

ft

To find the angle of depression, the best trigonometric ratio to use is the sine ratio. After equating it with the ratio of the opposite side over the hypotenuse, get the inverse of the sine ratio as shown:

sin

θ

=

o

p

p

h

y

p

θ

=

sin

−

1

(

o

p

p

h

y

p

)

θ

=

sin

−

1

(

1275

ft

5129

ft

)

θ

=

sin

−

1

(

0.2485864691

)

θ

=

14.3938824691

≈

14.4

∘

The angle of depression is

14.4

Answer: A someone with dangerous material passed by security

Step-by-step explanation:

Answer: -8, -3, 2, |-6|

Step-by-step explanation:

-3 , -8, |-6|, 2

|-6|=6

6>2

-3>-8 since a negative number closer to 0 is the greater number.

Order: -8, -3, 2, 6 ==> -8, -3, 2, |-6|

Find the critical points of [tex]f_1[/tex] and [tex]f_2[/tex].

By the fundamental theorem of calculus,

[tex]{f_1}'(x) = \displaystyle \prod_{j=1}^{21} (x-j)^j = (x-1) (x-2)^2 (x-3)^3 \cdots (x-21)^{21}[/tex]

and [tex]{f_1}'=0[/tex] for [tex]x\in\{1,2,3,\ldots,21\}[/tex]. The stationary points at odd values have odd multiplicity, while the even ones have even multiplicity. At the points of odd multiplicity, [tex]{f_1}'[/tex] passes through the [tex]x[/tex]-axis and the sign of [tex]{f_1}'[/tex] changes, while at the points of even multiplicity, [tex]{f_1}'[/tex] is tangent the to [tex]x[/tex]-axis. We also have

[tex]\displaystyle \lim_{x\to-\infty} {f_1}'(x) = (-1)^{1+3+5+\cdots+21} \infty = (-1)^{11^2} \infty = -\infty[/tex]

[tex]\displaystyle \lim_{x\to\infty} {f_1}'(x) = \infty[/tex]

so the plot of [tex]{f_1}'[/tex] looks more or less like the attached curve. (Not drawn to scale)

Wherever the sign of [tex]{f_1}'[/tex] changes from negative to positive, as it does in the case of [tex]x\in\{1,5,9,13,17,21\}[/tex], there is a local minimum, so [tex]m_1 = 6[/tex]. Similarly, wherever the sign changes from negative to positive, as it does when [tex]x\in\{3,7,11,15,19\}[/tex], there is a local maximum, so [tex]n_1 = 5[/tex].

Repeat for [tex]f_2[/tex].

[tex]{f_2}'(x) = 4900 (x-1)^{49} - 29400 (x-1)^{48} = 4900 (x-1)^{48} (x-7)[/tex]

has roots at [tex]x=1[/tex] and [tex]x=7[/tex], but only the latter is a critical point due to odd multiplicity. A rough sketch of the plot of [tex]{f_2}'[/tex] is also attached. The sign of [tex]{f_2}'[/tex] changes from negative to positive at [tex]x=7[/tex], so [tex]m_2 = 1[/tex] and [tex]n_2=0[/tex].

We conclude that

Q.9) [tex]2m_1+3n_1+m_1n_1 = \boxed{57}[/tex]

Q.10) [tex]6m_2+4n_2+8m_2n_2 = \boxed{6}[/tex]

Answer: $5.75

Step-by-step explanation: 2.75 x 2 = 5.5    1.25 x 3 = 2.75    2.75 + 5.5 = 9.25     $15 - $9.25 = $5.75

Answer:

USE AIR MATH!!

Step-by-step explanation:

Answer:

[tex]\textsf{1.} \quad x=-6, \quad x=6[/tex]

[tex]\textsf{2.} \quad x=-5, \quad x=2[/tex]

[tex]\textsf{3.} \quad x=-3, \quad x=7[/tex]

[tex]\textsf{4.} \quad x=\dfrac{-5 +\sqrt{57} }{4}, \quad x=\dfrac{-5 -\sqrt{57} }{4}[/tex]

[tex]\textsf{5.} \quad x=\dfrac{7+ \sqrt{309} }{10}, \quad x=\dfrac{7-\sqrt{309} }{10}[/tex]

Step-by-step explanation:

Method: Extracting the Square Root

[tex]\begin{aligned}& \textsf{Given}: & x^2 & = 36\\& \textsf{Square root both sides}: & \sqrt{x^2} & = \sqrt{36}\\& \textsf{Simplify}: & x & = \pm 6\\&\textsf{Solution}:&x& = -6, 6\end{aligned}[/tex]

Method: Factoring

[tex]\begin{aligned}& \textsf{Given}: & x^2+3x-10 & = 0\\& \textsf{Split the middle term}: & x^2+5x-2x-10 & = 0\\& \textsf{Factor the first two and the last two terms}: & x(x+5)-2(x+5)&=0\\& \textsf{Factor out the common term $(x+5)$}: & (x-2)(x+5)&=0\\& \textsf{Apply the zero-product property}: & (x-2)=0 \implies x&=2\\ &&(x+5)=0 \implies x&=-5\\ & \textsf{Solution}: & x&=-5,2\end{aligned}[/tex]

Method: Factoring

[tex]\begin{aligned}& \textsf{Given}: & x^2-4x-21 & = 0\\& \textsf{Split the middle term}: & x^2-7x+3x-21 & = 0\\& \textsf{Factor the first two and the last two terms}: & x(x-7)+3(x-7)&=0\\& \textsf{Factor out the common term $(x-7)$}: & (x+3)(x-7)&=0\\& \textsf{Apply the zero-product property}: & (x+3)=0 \implies x&=-3\\&&(x-7)=0 \implies x&=7\\& \textsf{Solution}: & x & = -3, 7\end{aligned}[/tex]

Method: Quadratic  Formula

[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}[/tex]

Given:

[tex]5x-4=-2x^2[/tex]

Rearrange into standard form by adding 2x² to both sides:

[tex]\implies 2x^2+5x-4=0[/tex]

Therefore:

[tex]a=2, \quad b=5, \quad c=-4[/tex]

Substitute the values into the quadratic formula and solve for x:

[tex]\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2)(-4)} }{2(2)}[/tex]

[tex]\implies x=\dfrac{-5 \pm \sqrt{25+32} }{4}[/tex]

[tex]\implies x=\dfrac{-5 \pm \sqrt{57} }{4}[/tex]

Therefore, the solutions are:

[tex]x=\dfrac{-5 +\sqrt{57} }{4}, \quad x=\dfrac{-5 -\sqrt{57} }{4}[/tex]

Method: Quadratic  Formula

[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}[/tex]

Given:

[tex]5x^2-7x-13=0[/tex]

Therefore:

[tex]a=5, \quad b=-7, \quad c=-13[/tex]

Substitute the values into the quadratic formula and solve for x:

[tex]\implies x=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(5)(-13)} }{2(5)}[/tex]

[tex]\implies x=\dfrac{7 \pm \sqrt{49+260} }{10}[/tex]

[tex]\implies x=\dfrac{7 \pm \sqrt{309} }{10}[/tex]

Therefore, the solutions are:

[tex]x=\dfrac{7+ \sqrt{309} }{10}, \quad x=\dfrac{7-\sqrt{309} }{10}[/tex]

2.5, 0.92, 0.75, 0.925