All categories

2 years ago

8. 6r + 5 = 4 6 - 7y -20

Mathematics

1 answer:

Rudiy272 years ago

3 0

Answer: r=-7/6y+7/2

Step-by-step explanation:

Step 1: Add -5 to both sides.

6r+5+−5=−7y+26+−5

6r=−7y+21

Step 2: Divide both sides by 6.

6r/6=-7y+21/6

You might be interested in

The correct answer is letter E, but I am not sure how to approach this problem and I was wondering if anyone can offer any guida

Jobisdone [24]

Explanation:

The key to area in polar coordinates is the formula for the area of a sector:

a = (1/2)r²θ

Then a differential of area* can be written as ...

da = (1/2)r²·dθ

Filling in the given function for r, we have ...

da = (1/2)(4cos(3θ))²·dθ = 8cos(3θ)²·dθ

The integral will have limits corresponding to the range of values of θ for one loop of the graph: -π/6 to π/6. So, the area is ...

$\displaystyle A=\int{da}=\int\limits_{-\pi /6}^{\pi /6}{8\cos{(3\theta)}^2}\, d\theta\\\\=8\int\limits_{-\pi /6}^{\pi /6}{\cos{(3\theta)}^2}$

_____

* As with other approaches to finding area (horizontal or vertical slice, for example), we assume that the differential element dθ is sufficiently small that we need not concern ourselves with the fact that r is a function of θ.

6 0

2 years ago

a circular dog pen has a circumference of 78.5 feet. approximate by 3.14 and estimate how many hunting dogs could be safely kept

olasank [31]

Answer:

Step-by-step explanation:

To solve this problem, we need to find the area of the pen, then divide that by the area required by each dog.

The relationship between circumference and radius is ...

C = 2πr

Solving for r gives ...

r = C/(2π)

The relationship between radius and area of a circle is ...

A = πr²

Substituting for r from above gives ...

A = π(C/(2π))² = C²/(4π)

Filling in the given numbers, we find the area of the pen to be ...

A = (78.5 ft)²/(4·3.14) = 490.625 ft²

Dividing the pen area by the area per dog gives the number of dogs the pen can hold:

(490.625 ft²)/(60 ft²/dog) ≈ 8.18 dogs

The pen can safely keep 8 dogs.

7 0

3 years ago

Select the correct answer from each drop-down menu.

ella [17]

Answer:

A non-equilateral rhombus.

Step-by-step explanation:

We can solve this graphically.

We start with square:

ABCD

with:

A = (11, - 7)

B = (9, - 4)

C = (11, - 1)

D = (13, - 4)

Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:

AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13

BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13

CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13

DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13

And we change C by C' = (11, 1)

In the image you can see the 5 points and the figure that they make:

The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.