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sdas [7]
4 years ago
6

Question 1

Mathematics
1 answer:
ASHA 777 [7]4 years ago
6 0
Question 1 Simplify the following expression: (5x2 + 3x + 4) − (2x2 − 6x + 3). If the final answer is written in the form Ax2 + Bx + C, what is the value of B?

Solution:

5x^2 + 3x + 4 - 2x^2 + 6x - 3 = 3x^2 + 9x + 1

Then, B = 9

Question 2
Simplify: (3x2 − 2) + (2x2 − 6x + 3)

5x2 − 6x + 1

5x2 − 6x − 1

x2− 6x + 1

5x2 − 8x + 3

Solution:

3x^2 - 2 + 2x^2 - 6x + 3 = 5x^2 - 6x + 1

Question 3
A sports company conducted a road test of a new model of a geared bike. The test rider cycled (3x - 2) miles on a flat road, (x2 - 5) miles uphill, and (2x + 7) miles downhill. Which simplified expression is equivalent to the total distance, in miles, for which the bike was tested?

x2 - x

x2 + 5x

x2 − x + 14

x2 + 5x + 14

(3x - 2) + (x2 - 5) + (2x + 7) = 3x - 2 + x^2 - 5 +2x + 7 = x^2 + 5x

Question 4
Simplify: (4x − 6) − (5x + 1)

x + 7

−x + 7

x − 7

−x − 7

(4x − 6) − (5x + 1) = 4x - 6 - 5x - 1 = -x -7

Question 5
Simplify the following expression: (x + 6y) - (3x − 10y). If the final answer is written in the form Ax + By, what is the value of A?

(x + 6y) - (3x − 10y) = x + 6y - 3x + 10y = -2x + 16y

A = -2

Question 6
Simplify: (3x − 5) + (3x + 6)

6x − 1

1

6x − 11

(3x − 5) + (3x + 6) = 3x - 5 + 3x + 6 = 6x + 1


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a. If cosθ=−45 where θ is in quadrant 3, find sin2θ. b. If cosθ=2√2 where θ is in quadrant 1, find cos2θ. c. If sinθ=817 where θ
sesenic [268]

Answer:

Part A) sin(2\theta)=\frac{24}{25}

Part B) cos(2\theta)=0

Part C) tan(2\theta)=-\frac{240}{161}

Step-by-step explanation:

Part A) we have cos(\theta)=-\frac{4}{5}

θ is in quadrant 3 ----> the sine is negative

Find sin(2\theta)

we know that

sin(2\theta)=2sin(\theta)cos(\theta)

Remember that

cos^{2} (\theta)+sin^{2} (\theta)=1

substitute

(-\frac{4}{5})^{2}+sin^{2} (\theta)=1

(\frac{16}{25})+sin^{2} (\theta)=1

sin^{2} (\theta)=1-\frac{16}{25}

sin^{2} (\theta)=\frac{9}{25}

sin(\theta)=-\frac{3}{5} ---> remember that the sine is negative (3 quadrant)

Find sin(2\theta)

we have

cos(\theta)=-\frac{4}{5}

sin(\theta)=-\frac{3}{5}

sin(2\theta)=2sin(\theta)cos(\theta)

substitute

sin(2\theta)=2(-\frac{3}{5})(-\frac{4}{5})

sin(2\theta)=\frac{24}{25}

Part B) we have cos(\theta)=\frac{\sqrt{2}}{2}

θ is in quadrant 1

Find cos(2\theta)      

we know that

cos(2\theta)=2cos^{2} (\theta)-1

substitute

cos(2\theta)=2(\frac{\sqrt{2}}{2} )^{2}-1

cos(2\theta)=0

Part C) we have sin(\theta)=\frac{8}{17}

θ is in quadrant 2 ----> the cosine is negative

Find tan(2\theta)  

we know that

tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}

Remember that

cos^{2} (\theta)+sin^{2} (\theta)=1

substitute

cos^{2} (\theta)+(\frac{8}{17})^{2}=1

cos^{2} (\theta)=1-\frac{64}{289}

cos^{2} (\theta)=\frac{225}{289}

cos(\theta)=-\frac{15}{17}

Find tan(\theta)  

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}

substitute

tan(\theta)=\frac{(8/17)}{(-15/17)}

tan(\theta)=-\frac{8}{15}

Find tan(2\theta)  

tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}

substitute

tan(2\theta)=\frac{2(-\frac{8}{15})}{1-(-\frac{8}{15})^{2}}

tan(2\theta)=\frac{(-\frac{16}{15})}{1-(\frac{64}{225})}

tan(2\theta)=\frac{(-\frac{16}{15})}{1-\frac{64}{225}}

tan(2\theta)=\frac{(-\frac{16}{15})}{\frac{161}{225}}

tan(2\theta)=-\frac{240}{161}

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Step-by-step explanation:

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