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2 years ago

Point O is the center of the circle.

Mathematics

1 answer:

photoshop1234 [79]2 years ago

6 0

Answer:

Step-by-step explanation:

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Help please asap PLEASE TELL ME THE WORKING AS WELL.

DanielleElmas [232]

Answer:

a = 9 / 7

step by step explanation:

9 / - a + 15 = 8

Determined the defined range

9 / ( - a ) + 15 = 8 , a ≠ 0

subtract 15 from each side

9 / - a + 15 - 15 = 8 - 15

calculate the difference

9 / - a = - 7

multiply both side of equation by - a

9 / - a × - a = - 7 × - a

9 = 7 a

divide each side by of equation by -7

9 / 7 = 7 a / 7

9 / 7 = a

swap the side of equation . ( optional )

a = 9 / 7

6 0

2 years ago

Read 2 more answers

Simplify the expression. (7x + 3) + (4x - 1) - (2x + 4)

Crazy boy [7]

Answer:

9x -2

Step-by-step explanation:

(7x + 3) + (4x - 1) - (2x + 4)

Distribute the minus sign

(7x + 3) + (4x - 1) - 2x - 4

Combine like terms

7x+4x-2x +3 -1-4

9x -2

3 0

2 years ago

What is the domain and range of relation (4,3)(-2,2)(5,-6)

vazorg [7]

Domain: {-2, 4, 5}
Range {-6, 2, 3}

5 0

2 years ago

Please help me for number 1 and number 3

azamat

The middle is 2 :D hope this helps

8 0

3 years ago

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

3 years ago