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ki77a [65]
3 years ago
13

Which table contains ordered pairs that lie on the graph of the equation 5x + 3y = 15?

Mathematics
2 answers:
Alenkinab [10]3 years ago
5 0

Answer:

D is the right answer

5(0) + 3(5) =15

5(3) + 3(0) =15

Step-by-step explanation:

topjm [15]3 years ago
3 0

Answer:

5x + 3y = 15

x- int y= 0

5x + 3(0) = 15

5x = 15

x = 3

(3,0)

y-int x=0

5(0) + 3y = 15

3y = 15

y =5

(0,5)

The solution is d

Step-by-step explanation:

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Answer:

C. Correct

Step-by-step explanation:

The interpretation of a confidence interval of x% is that we are x% that this interval contains the mean of the population.

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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]\\

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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Answer:

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