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den301095 [7]
2 years ago
8

Jamal has $15.00 to spend at the concession stand. He buys nachos for $7.50, and he wants to purchase some sour straws for $1.50

each. How many sour straws can Jamal purchase with the money that he has?
Mathematics
1 answer:
frosja888 [35]2 years ago
5 0
15-7.50=7.50
7.50/1.50=?
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Polygon ABCD is reflected and dilated to give polygon PQRS. The coordinates of the preimage are (2, 2), (6, 8), (12, 8), and (16
HACTEHA [7]

Answer: The scale factor of the dilation is 0.5.

Explanation:

It is given that ABCD is polygon which is reflected and dilated , then we get PQRS.

The vertices of ABCD are (2, 2), (6, 8), (12, 8), and (16, 2) respectively. The vertices of image  PQRS are (11, 15),(9, 12), (6, 12), and (4, 15) respectively.

The reflection affects the coordinates but does not affect the length of the sides.

But when we talk about dilation it affects the length of sides according to the scale factor or a constant factor.

If a line segment AB is dilated by scale factor k then length of A'B' is k times length of AB.

|A'B'|=k\times |AB|   .... (1)

So we have to find any side length of preimage and the side length of that side in image. it means we have to find AB and PQ.

Use distance formula to find the side length.

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AB=\sqrt{(8-2)^2+(6-2)^2}=\sqrt{36+16}=2\sqrt{13}

PQ=\sqrt{(9-11)^2+(12-15)^2}=\sqrt{4+9}=\sqrt{13}

It is noticed that the size of side in preimage is 2\sqrt{13} and the size of same side in image is \sqrt{13}.

Using equation (1), we get

\sqrt{13}=k \times 2\sqrt{13}

k=\frac{1}{2}

k=0.5

Hence, the value of scale factor is 0.5.

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Enter an equation that gives the cost y of renting a car for x days from Rent-All.
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Answer:At the time of rental, an authorized hold will be secured on your credit/debit card provided to cover the estimated rental charges plus an additional $200.00 to cover additional charges that may be incurred.

Step-by-step explanation:

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Factor 18x^3+24x^2-21x-28
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Hello my name is Jeff (sorry)
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3 years ago
Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

6 0
1 year ago
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Answer:

4(x+3)

Step-by-step explanation:

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