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3 years ago

8

Hi! 8th grade math problem for you. Please see the screenshot

1 answer:

deff fn [24]3 years ago

4 0

Answer:

r=4/2=2cm

total volume=volume of cylinder + cone=πr²h+1/3πr²h'=πr²(h+1/3 h')=2²π((10-3)+1/3× 3)=4π(7+1)=8×4π=32πcm³

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Help please please <br> $)

mart [117]

<h3>Look at the picture.</h3>

<em>commutative property</em> a + b = b + a

... (-5 + 1/2 + 5) = ... (-5 + 5 + 1/2)

<em>inverse property of addition</em> a + (-a) = a - a = 0

... (-5 + 5 + 1/2) = ... (0 + 1/2)

<em>identity property of addition</em> a + 0 = 0 + a = a

... (0 + 1/2) = ... 1/2

<em>inverse property of multiplication</em> a/b · b/a = 1

37 · 2 · 1/2 = 37 · 2/1 · 1/2 = 37 · 1

<em>identity property of multiplication</em> a · 1 = 1 · a = a

37 · 1 = 37

6 0

3 years ago

Someone please help me

riadik2000 [5.3K]

Answer:

5.625

Step-by-step explanation:

So first do 112 / 360

Simplify it = 28 / 90 = 14 / 45

So 1.75 is 14 / 45 of the circumference.

The whole circumference is 1.75 * (45 / 14)

So the answer is 45 / 8 = 5.625

3 0

3 years ago

Find the LCM of 2 and 21

Alexus [3.1K]

The answer is 10 because it had

6 0

4 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

3 years ago

PLEASE HELP ASAP!!! WILL GIVE BRAINLIEST FOR BEST/RIGHT ANSWER!!!!!!!!

Nesterboy [21]

The answer is A, please mark brainliest.

7 0

3 years ago

Read 2 more answers