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Tems11 [23]
3 years ago
5

Find the volume of a cone with a radius of 8 cm and height 15 cm? Round to the nearest tenth. Please show work.

Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0
You multiply radius by 3.14 by the height  if you plug it in your equation is 8×3.14×15 which equals 1005.31 which rounds to 1005, the answer is 1005! :D
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A rectangular shaped pond is 16 meters long and 14 meters wide. The dimensions of a rectangular shaped lake nearby is 350%
aalyn [17]

Answer: 56 meters long and 49 meters wide

Step-by-step explanation:

Convert 350% to a decimal. 350% is 3.5. Each lenth is 3.5x larger than the original pond.

16*(3.5) =56 meters length

14*(3.5) = 49 meters width

56 meters long and 49 meters wide

4 0
3 years ago
Suppose f(x, y) is a differentiable function of x and y and let g(r, s) = f (2rs, 8s − 2r). Use
worty [1.4K]

By the chain rule,

\dfrac{\partial g}{\partial r} = \dfrac{\partial f}{\partial r} \\\\ ~~~~~ = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial r} \\\\ ~~~~~ = 2 s \dfrac{\partial f}{\partial x} - 2 \dfrac{\partial f}{\partial y}

where x(r,s)=2rs and y(r,s)=8s-2r.

If r=2 and s=1, then

x = 2\cdot2\cdot1 = 4

y=8\cdot1-2\cdot2 = 4

so that

g_r(2,1) = 2\cdot1 f_x(4,4) - 2 f_y(4,4) = 2\cdot2-2\cdot3 = \boxed{-2}

5 0
1 year ago
I NEED HELP PLEASE, THANKS
castortr0y [4]

First, find the scale factor.

8.4 / 7 = 1.2

9 / 7.5 = 1.2

7.2 / 6 = 1.2

Since both solids are different sizes, the solids aren't congruent.

The size ratio for all the sides are:

8.4:7

9:7.5

7.2:6

Since the scale factor isn't 1:1 this also proves that the solids are NOT congruent.

Since both solids are the same kind of shape and have an identical scale factor, the solids are similar.

Best of Luck!

5 0
3 years ago
Write the standard equation for the circle center (-6,8) that passes through (0,0)
Reil [10]

Answer:

D

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (- 6, - 8), thus

(x + 6)² + (y + 8)² = r²

The radius is the distance from the centre to a point on the circle

Calculate r using the distance formula

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 6, - 8) and (x₂, y₂ ) = (0, 0)

r = \sqrt{(0+6)^2+(0+8)^2} = \sqrt{36+64} = \sqrt{100} = 10

Hence

(x + 6)² + (y + 8)² = 100 → D

4 0
3 years ago
Determine whether the statement is always true, sometimes true, or never true. Give an example. (a) Both sides of an equation ca
arsen [322]

Answer:

(A) Always true

(B) Seldom true

Step-by-step explanation:

(A) Both sides of an equation can be multiplied by the same number without this action changing the solution of the equation.

Keyword here is "number".

Multiplying both sides of an equation by an equal quantity (same number) will result in same solution after solving for the unknown in the equation. This is because that numeric quantity can always be removed by dividing both sides of the equation by it or by a factor (or multiple) of it.

EXAMPLE:

For a linear equation, 4x - 6 = 15x

let's find the solution, in other words solve for the unknown value X.

4x - 15x = 6

-11x = 6

x = -6/11

Now multiply both sides of the equation by 3.

3(4x - 6) = 3(15x)

12x - 18 = 45x   ___new equation

Solve for X

12x - 45x = 18

-33x = 18

x = -18/33

Reduce the fraction to its simplest form by looking for a number that can divide both numerator and denominator without remainder. In other words, think of a number that is a factor of 18 and a factor of 33.

That common factor or highest common factor (HCF) is 3.

Go ahead and reduce the fraction.

x will be reduced to -6/11

(B) Both sides of an inequality can seldom be multiplied by the same number, without such action changing the solution set of the equation.

Inequalities are more complex. Operational signs even change sometimes, in the course of finding the solution set of the inequality.

Sometimes, multiplying both sides of an inequality by a given numeric quantity will change its solution set and sometimes, it won't.

5 0
3 years ago
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