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Leviafan [203]
3 years ago
6

10-11d>-5d-4 show step by step

Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0
The first step for solving this is to move the variable to the left side and then change its sign.
10 - 11d + 5d > - 4
Now move the constant to the right side and change its sign.
-11d + 5d > -4 - 10
Collect the terms with a d variable.
-6d > -4 - 10
Calculate the difference on the right side.
-6d > -14
Lastly,, divide both sides of the inequality by -6 and flip the inequality sign to find your final answer.
d < \frac{7}{3}
Let me know if you have any further questions.
:)
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Classify the pair of angles. Then find the value of x.
sleet_krkn [62]

Answer:

3) x = 50, vertical angles.

4) x = 20, supplementary angles

5) x = 14, complimentary angles

Step-by-step explanation:

<h3>3)</h3>

<em>vertical angles.</em>

<em>vertical angles are equal.</em>

<em>so, you can make this equation:</em>

67 = x + 17

<em>subtract 17 from both sides of the equation</em>

67 - 17 = x + 17 - 17

67 - 17 = x

50 = x

<h3>4)</h3>

<em>supplementary angles</em>

<em>supplementary angles add up to 180</em>

<em>so, you can make this equation:</em>

7x + 40 = 180

<em>subtract 40 from both sides of the equation</em>

7x + 40 - 40 = 180 - 40

7x = 180 - 40

7x = 140

<em>divide both sides by 7</em>

7x/7 = 140/7

x = 140/7

x = 20

<h3>5)</h3>

<em>complimentary angles</em>

<em>complimentary angles add up to 90</em>

<em>so, you can make this equation:</em>

4x + 34 = 90

<em>subtract 34 from both sides of the equation</em>

4x + 34 - 34 = 90 - 34

4x = 90 - 34

4x = 56

<em>divide both sides by 4</em>

4x/4 = 56/4

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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]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

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