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2 years ago

15 points PLEASE HELP , and be careful when answering !!

Mathematics

1 answer:

Aleks04 [339]2 years ago

6 0

18.84 because the formula to find the volume of a cone is pie*r^2*h/3 and your radius is 3 and the height is 2, so if you plug those in the equation and calculate it this should be your answer

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Solve the inequality below: <br><br> -2(x-5)<4

alexandr402 [8]

Answer:

x > 3

Step-by-step explanation:

-2(x-5) < 4

-2x+10 < 4

-2x < -6

x > 3

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2 years ago

Where are all the math whizzes? I need your help, please! I will mark brainliest!

hjlf

Answer:

Step-by-step explanation:

A. x/yx * zy/yx = 2/x

B. x/y + y/z = (xz+y^2)/yz

C. x*z/z*x = 1

D. (x+y)/(y+z) = y(x+1)/y(1+z) = x+1/1+z

E. x/y / z/y = x/y * y/z = x/z

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2 years ago

On the grid below, which point is located in the quadrant where the x-coordinate is a positive number and the y-coordinate is a

leonid [27]

Answer:

Step-by-step explanation:

The answer is C.

5 0

2 years ago

D -e/3d + e= e, for d

navik [9.2K]

Answer:

$\large\boxed{d=\dfrac{e^2+e}{1-3e}}$

Step-by-step explanation:

$\dfrac{d-e}{3d+e}=e\\\\\dfrac{d-e}{3d+e}=\dfrac{e}{1}\qquad\text{cross multiply}\\\\(1)(d-e)=(e)(3d+e)\qquad\text{use the distributive property}\\\\d-e=3de+e^2\qquad\text{add}\ e\ \text{to both sides}\\\\d=3de+e+e^2\qquad\text{subtract}\ 3de\ \text{from both sides}\\\\d-3de=e^2+e\qquad\text{distribute}\\\\d(1-3e)=e^2+e\qquad\text{divide both sides by}\ (1-3e)\\\\d=\dfrac{e^2+e}{1-3e}$

8 0

2 years ago

PLEASE HELP ILL GIVE BRAINLIEST

dusya [7]

$(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}$

let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

$\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}$

7 0

2 years ago

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