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Mariana [72]
3 years ago
11

2y - A / 3 equals a + 3 B / 4 solving for y

Mathematics
1 answer:
melamori03 [73]3 years ago
8 0
To solve for y you need to first multiply both sides by 4. This will give you: 4y - 4A / 3 = A + 3B

Next you need to multiply both sides by 3: 4y - 4A = 3A + 9B

Now you need to add 4A to both sides: 4y = 7A + 9B

Now divide both sides by 4:

Y = 7A/4 + 9B/4
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Combine like terms for <br> (3x²+2x²-5x+7)+(4x²+2x-3)
horsena [70]

Answer:

9x^2-3x+4

Step-by-step explanation:

(3x^2+2x^2-5x+7)+(4x^2+2x-3)

5x^2-5x+7+4x^2+2x-3

5x^2+4x^2-5x+2x+7-3

9x^2-3x+7-3

9x^2-3x+4

5 0
3 years ago
Read 2 more answers
Just the last two ones please and thank you if you awnsered this
Rudik [331]
Number 1 is twenty and number two is fifty six
8 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
I need help with this one <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20" id="TexFormula1" title=" \frac{1}{3} "
Vanyuwa [196]
To solve this questions, we can turn 2 into a fractions.

2 as a fraction is 2/1 (because 2 divide by 1 is still 2)
We can now use this to get our answer. 

1/3  ÷ 2/1 
=  1/3 × 1/2    (flip the fraction and then multiply)
= 1/6          (1 x 1 / 3 x 6)

Or 0.16666.. as a decimal
7 0
3 years ago
Elijah buys a one-pint bottle of juice for $2.88. What is the unit rate of the cost of the juice per fluid ounce?
mojhsa [17]
Answer: 0.48
Explanation: $2.88 / 16
3 0
3 years ago
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