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3 years ago

What is the value of x

Mathematics

1 answer:

pochemuha3 years ago

5 0

This is the isosceles triangle. Therefore, the angles at the base are congruent.

We know that the sum of the measures of the angles in a triangle is 180°.

Therefore we have the equation:

$38^o+x^o+x^o=180^o$

$38^o+2x^o=180^o$ <em>subtract 38 from both sides</em>

$2x^o=142^o$ <em>divide both sides by 2</em>

$x^2=71^o$

<h3>Answer: x = 71.</h3>

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The lengths in feet of three pieces of timber are 48, 72, and 40. The sawmill operator needs to cut the timber into logs of equa

olya-2409 [2.1K]

<h3>Therefore she can cut 8 feet long log.</h3>

Step-by-step explanation:

Given,

The lengths in feet of three pieces of timber are 48,72 and 40. The sawmill operator needs to cut the timber into logs of equal length with no waste.

The HCF of 48 , 72 and 40 is =8

Therefore she can cut 8 feet long log.

3 0

3 years ago

What steps do I need to do to get 18×8.7<br><br>23×56.1<br><br>and<br><br>47×5.92

eimsori [14]

For 18 x 8.7 = 156.6

For 23 x 56.1 = 1290.3

For 47 x 5.92 = 278.24

but i don't know how to explain the steps, just the answers.
but hope i helped.

3 0

3 years ago

Read 2 more answers

What is the image of (-8,6) after a translation over the line y=-x

stira [4]

Answer:

(-6, 8)

Step-by-step explanation:

(x, y) --> (-y, -x)

(-8, 6) --> (-6, 8)

Therefore, (-6, 8)

I hope this helped and have a good rest of your day!

8 0

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

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

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