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3 years ago

10

find the sum ofthe angle measure in each regular polygon. Then find the measure of each angle. Show work please I'm so bad at th

is

1 answer:

Tems11 [23]3 years ago

6 0

For every side you just add 90 to your equation. so #4 would be 360-125-125-55=(55)

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I don't understand this problem can someone please help me

Stella [2.4K]

ok kid you need to study instead of getting people to give you the answer

5 0

3 years ago

ILL GIVE BRAINLIEST<br><br> evaluate the expression. 2•7(3^2)/3

scoray [572]

Answer:

$\frac{7 \times {3}^{2} }{3} \\ \frac{7 \times 3 \times 3}{3} \\ = 7 \times 3 = 21 \\ thank \: you$

7 0

3 years ago

Read 2 more answers

Hey can you please help me posted picture of question

Maksim231197 [3]

For this case we have the following equation:
y = x2-4x + 3
Deriving we have the following equation:
y '= 2x-4
We equal zero and clear x:
2x-4 = 0
x = 4/2
x = 2
Substituting in the given equation we have:
y = (2) ^ 2-4 (2) +3
y = 4-8 + 3
y = -1
The vertex will be the ordered pair:
(x, y) = (2, -1)
Answer:
(x, y) = (2, -1)
option B

5 0

3 years ago

Read 2 more answers

Anyone mind helping!?

Ksju [112]

Answer:

B) $3\sqrt{3}$

Step-by-step explanation:

First simplify the radical on the right.

$\sqrt{75}$ is equal to $\sqrt{25} *\sqrt{3}$ , because 25 * 3 is 75, so we can separate the square roots into the product of these two square roots.

Then we know $\sqrt{25}$ is just equal to 25, which means the simplified version of $\sqrt{75}$ is $5\sqrt{3}$ .

Then we can use this new representation to finish solving.

$8\sqrt{3}-5\sqrt{3} =3\sqrt{3}$ , so the answer would be B

6 0

3 years ago

Read 2 more answers

a spherical balloon is deflated at a rate of 256pi/3 cm^3/sec. at what rate is the radius of the balloon changing when the radiu

Hunter-Best [27]

What you need to look for is $\frac { dr }{ dt }$ when r=8.

Now:

$Volume\quad of\quad a\quad sphere:\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }$

$\therefore \quad \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi \cdot 3{ r }^{ 2 }=4\pi { r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dV } =\frac { 1 }{ \frac { dV }{ dr } } =\frac { 1 }{ 4\pi { r }^{ 2 } }$

$And:\\ \\ \frac { dV }{ dt } =-\frac { 256\pi }{ 3 } \\ \\ Therefore:\\ \\ \frac { dr }{ dt } =\frac { dr }{ dV } \cdot \frac { dV }{ dt }$

$\\ \\ =\frac { 1 }{ 4\pi { r }^{ 2 } } \cdot -\frac { 256\pi }{ 3 } \\ \\ =-\frac { 256 }{ 12 } \cdot \frac { \pi }{ \pi } \cdot \frac { 1 }{ { r }^{ 2 } }$

$\\ \\ =-\frac { 64 }{ 3{ r }^{ 2 } } \\ \\ When\quad r=8,\\ \\ \frac { dr }{ dt } =-\frac { 64 }{ 3\cdot { 8 }^{ 2 } } =-\frac { 1 }{ 3 } \\ \\ Answer:\quad -\frac { 1 }{ 3 } \quad cm/sec$

3 0

4 years ago