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3 years ago

Any help please????Anyone

Mathematics

1 answer:

Svet_ta [14]3 years ago

7 0

<ACF = <FGA

<GAC = <CFG

<GAC + <ACF = 180°

136° + <ACF = 180°

<ACF = 180° - 136°

<ACF = 44°

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Thepotemich [5.8K]

Answer:

Step-by-step explanation:

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3 years ago

A right rectangle prism has a length of 5 cm, a width of 4 cm, and a height of of 3 cm. The dimensions of the prism are doubled.

MaRussiya [10]

As the volume of cuboid is given by,
l×b×h
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3 years ago

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Ely has a bag of 42 strawberries. She eats 16 strawberries. How many cherries does she have left. The next day she goes to the s

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The number of cherries left is 26

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<em><u>Solution:</u></em>

Given that, Ely has a bag of 42 strawberries. She eats 16 strawberries

To find: Number of cherries left

From given,

Total number of strawberries = 42

Number of strawberries eaten = 16

Therefore,

Number of cherries left = Total number of strawberries - Number of strawberries eaten

Number of cherries left = 42 - 16 = 26

Thus, number of cherries left is 26

<em><u>The next day she goes to the store and buys 28 more cherries</u></em>

Number of cherries bought = 28

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Total number of cherries she has now = number of cherries left + Number of cherries bought

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3 years ago

Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places

natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

$\rm 5 \sin \theta = 4\\\\\theta = 0.927 , 2.214$

Then by the integration, we have

$\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\$

$\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\$