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3 years ago

6

Pls help me with this problem. Also the answer has to be in terms of pi

1 answer:

katovenus [111]3 years ago

4 0

Answer: 25pi
Explanation: area = piR^2 so 10(radius)^2 = 100pi, only 1/4 of the circle is there so divide the area found (100pi) by 4 to find the one quarter of the circle.

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If y varies directly as x and k = 8, find x when y = 8<br><br> 1. 1<br> 2. 8<br> 3. 64

aksik [14]

y = kx so x = y/k

if y = 8 and k = 8 then

x = 8/8 = 1

answer

1. 1

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3 years ago

azamat

Answer:

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Step-by-step explanation:

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3 years ago

Solve the equation for P: 3k=7Q+6p

iren [92.7K]

3k=7Q+6p

Solve for P which means we have to make 'p' alone

$3k=7Q+6p\\\\ Add 7Q on both sides \\\\ 3k - 7Q = 7Q - 7Q + 6p\\\\ \left ( 3k - 7Q \right ) = 6p\\\\ Divide both side by 6 to make p alone \\\\ \frac{\left ( 3k - 7Q \right )}{6} = \frac{6p}{6}\\\\ p = \frac{\left ( 3k - 7Q \right )}{6}$

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3 years ago

Read 2 more answers

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

7 0

3 years ago

Read 2 more answers

6p-10c-8 when p=6 and c=8

ELEN [110]

The answer is 36 in simplist form

5 0

3 years ago