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3 years ago

Point A is located at (3, 4) and is rotated 90° counterclockwise about the origin. The new location, point A’, is (-4, 3).

Mathematics

1 answer:

Nana76 [90]3 years ago

8 0

True. The 90° counterclockwise rule is (x,y) -> (-y,x)

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When p = 12, t = 2, and s = 1/6, r = 18. If r varies directly with p and inversely with the product of s and t, what is the cons

Ilia_Sergeevich [38]

R = kp/st
18 = 12k/(1/6 x 2)
18 = 12k/(1/3)
18 = 36k
k = 18/36 = 1/2

3 0

3 years ago

Polynomial add or subtract (-2x^2-4x+13)+(12x^2+2x-25)

BigorU [14]

1.Simplify.

-2{x}^{2}-4x+13+12{x}^{2}+2x-25−2x2−4x+13+12x2+2x−25

2.Collect like terms.

(-2{x}^{2}+12{x}^{2})+(-4x+2x)+(13-25)(−2x2+12x2)+(−4x+2x)+(13−25)

3.Simplify.

10{x}^{2}-2x-1210x2−2x−12

6 0

3 years ago

Urgent! Will award brainliest... Factor the following polynomial: <img src="https://tex.z-dn.net/?f=3a%5E%7B2%7D-21a%2B3

umka2103 [35]

Answer:

$3 (a-2) (a-5)$

Step-by-step explanation:

The first step is to find the GCF. Here, it's 3.

$3(a^2-7a+10)$

Then, you factor the polynomial in the parenthesis.

To find the factors, you will need to find 2 numbers that add to -7, and multiply to 10. -2 and -5 add to -7 and multiply to 10. Now, replace -7a with the factors.

$a^2-2a-5a+10$

This of this polynomial as 2 problems.

$a^2-2a$ $-5a+10$

Then, factor again.

$a^2-2a$ $-5a+10$

$a(a-2)$ $-5(a-5)$

Then, you keep the factors in parenthesis, and combine the numbers on the outside.

$(a-2)(a-5)(a-5)$

Since, there are 2 of the same factor, you only need one.

$(a-2)(a-5)$

BUT REMEMBER!! In the very beginning, we had a 3 that we took out, we STILL need to add that to the final answer. The final answer is:

$3 (a-2) (a-5)$

7 0

3 years ago

Which of the five measures of center (the mean, the median, the trimmed mean, the weighted mean, and the mode) can be calculated

nalin [4]

The mean and median can be calculated for quantitative data only.

4 0

3 years ago

A storage tank will have a circular base of radius r and a height of r. The tank can be either cylindrical or hemispherical (ha

Neko [114]

Answer:

Volume: $\frac{2}{3}\pi r^3$

Ratio: $\frac{2}{9}r$

Step-by-step explanation:

First of all, we need to find the volume of the hemispherical tank.

The volume of a sphere is given by:

$V=\frac{4}{3}\pi r^3$

where

r is the radius of the sphere

V is the volume

Here, we have a hemispherical tank: a hemisphere is exactly a sphere cut in a half, so its volume is half that of the sphere:

$V'=\frac{V}{2}=\frac{\frac{4}{3}\pi r^3}{2}=\frac{2}{3}\pi r^3$

Now we want to find the ratio between the volume of the hemisphere and its surface area.

The surface area of a sphere is

$A=4 \pi r^2$

For a hemisphere, the area of the curved part of the surface is therefore half of this value, so $2\pi r^2$ . Moreover, we have to add the surface of the base, which is $\pi r^2$ . So the total surface area of the hemispherical tank is

$A'=2\pi r^2 + \pi r^2 = 3 \pi r^2$

Therefore, the ratio betwen the volume and the surface area of the hemisphere is

$\frac{V'}{A'}=\frac{\frac{2}{3}\pi r^3}{3\pi r^2}=\frac{2}{9}r$

6 0

3 years ago