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3 years ago

PLEASE PLEASE HELP AND PLEASE DON'T GUESS AND SHOW YOUR WORK THANK YOU!!!!!!!!!

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

4 0

Would it be distribution were you multiply the three in? -3(b-7) -3*b and -3*-7 where your answer would be -3b+21

4vir4ik [10]3 years ago

3 0

The answer would be -3-7

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Y=4x + 3 and 2x + y = 39 how can I find the answer

Sunny_sXe [5.5K]

Answer:

x=6

Step-by-step explanation:

2x+(4x+3)=39

2x+4x+3=39

6x+3=39

6x=39-3

6x=36

x=36÷6

x=6

3 0

3 years ago

Please hurry I will mark you brainliest

Irina18 [472]

Answer:

The solution to above problem is 1- $45n 2- $(250+28n) 3- $(500+20n)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

PLEASE HELP 20 POINTS<br> solve the system of equations y=-5x-7 and -4x-3y=-1 by substitution.

Ainat [17]

Answer:

Solution: x = -2; y = 3 or (-2, 3)

Step-by-step explanation:

<u>Equation 1:</u> y = -5x - 7

<u>Equation 2:</u> -4x - 3y = -1

Substitute the value of y in Equation 1 into the Equation 2:

-4x - 3(-5x - 7) = -1

-4x +15x + 21 = -1

Combine like terms:

11x + 21 = - 1

Subtract 21 from both sides:

11x + 21 - 21 = - 1 - 21

11x = -22

Divide both sides by 11 to solve for x:

11x/11 = -22/11

x = -2

Now that we have the value for x, substitute x = 2 into Equation 2 to solve for y:

-4x - 3y = -1

-4(-2) - 3y = -1

8 - 3y = -1

Subtract 8 from both sides:

8 - 8 - 3y = -1 - 8

-3y = -9

Divide both sides by -3 to solve for y:

-3y/-3 = -9/-3

y = 3

Therefore, the solution to the given systems of linear equations is:

x = -2; y = 3 or (-2, 3)

Please mark my answers as the Brainliest if you find this helpful :)

7 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

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

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