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3 years ago

Ratios that are equivalent to 7: 6.

Mathematics

1 answer:

Alex3 years ago

7 0

Step-by-step explanation:

21:18=7:6

42:36=7:6

63:54=7:6

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Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

4 years ago

ONE MORE AFTER THIS PLEASE HELP ME

adoni [48]

the answer ur looking for is

y + 3

4 0

3 years ago

Read 2 more answers

Does the point (6,3) satisfy the equation y = 4(x – 5)?

Darina [25.2K]

Answer:

Step-by-step explanation:

plug (6,3) into the given equation to find out

3 = 4(6-5)

3 = 4(1)

3 = 4

since 3 does not equal 4, the point (6,3) does not satisfy the equation y=4(x-5)

5 0

3 years ago

An injectable solution contains 500mg in each 0.7 mL<br><br> Express above as a ratio.

Elden [556K]

650mg per 1 mL, I'm not completely sure of the answer, as you did not provide enough information.<span />

4 0

3 years ago

7. If f(x) = -x + 7, what is f(15)?Of(15) = 22f(15) = -22Of(15) = -8f(15) = 8

zysi [14]

the fuction f(x) is equal to:

$f(x)=-x+7$

now, to evaluate f(15) we have to replace x=15 so the equation is going to be equal to:

$f(15)=-15+7=-8$

6 0

1 year ago