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3 years ago

How do I solve for the variables r,s, and t

Mathematics

1 answer:

Softa [21]3 years ago

4 0

When using variables, and the whole question is full of variables. The letters can stand for anything.

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If y varies directly as x, and y is 18 when x is 5, which expression can be used to find the value of y when x is 11?

Rzqust [24]

The answer would be choice B y=18/5(11)

5 0

3 years ago

Peter and his four brothers combined all of their money to buy a video game. If 30% of the total money is Peter's, and $12.00 of

kari74 [83]

Answer:

$20.00

Step-by-step explanation:

t represents the total amount

0.25t = 5.00

Can you solve for t and answer?

Or logically

Peter's $5.00 is 1/4 (25%) of the total.

Therefore 4/4 is $20.00

6 0

2 years ago

Read 2 more answers

The radius of a circle is 13 centimeters. What is the circle's circumference? Use 3.14 for .

vlabodo [156]

Answer:

The circumference is 81.64cm

Step-by-step explanation:

Circumference of a circle=2×3.14×radius

2×3.14×13=81.64

I hope this helps you in any way :)

8 0

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.

Using Pythagoras Theorem

(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)

First derivative

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

apply chain rule

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

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

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

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

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

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

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

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

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

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

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers

WHO LIVES IN A PINNAPLE UNDER DA SEA ...?

Gala2k [10]

Answer:

patrick star

Step-by-step explanation:

5 0

3 years ago