Ask question

Ask question

All categories

jek_recluse [69]

3 years ago

10

Find the area in the first quadrant bounded by the arc of the circle described by the polar equation r = 2 sin θ + 4 cos θ. The

circle is graphed in the accompanying figure. A. 5π/2 + 4 B. 5 π C. 5 π + 8 D. 5π/2

1 answer:

dedylja [7]3 years ago

6 0

The radius of the circle is sqrt(5)

so the A = pi*r*r = pi * sqrt(5) * sqrt(5) = 5pi

the answer os <span>B. 5 π </span>

You might be interested in

PLEASE HELP DUE IN 3 minutes

son4ous [18]

Answer:

2....................

3 0

2 years ago

Read 2 more answers

PLEASE HELP!!!!!!!! WILL MARK BRAINLIEST!!!! aLSO 99 POINTS!!!!!!!! If a train is scheduled to go 87 and 1 half miles in 2 and 1

Roman55 [17]

The answer is C. 35 mph

3 0

2 years ago

Read 2 more answers

Please help asap !!!!!! what vaule of x makes the equation true 3(x-4)=15

DaniilM [7]

3(x-4) = 15

Distribute the 3:

3x-12 = 15

Add 12 to both sides:

3x = 27

Divide both sides by 3:

X = 9

The answer is D. 9

5 0

2 years ago

Terry can run 15 miles in the same amount of time it takes mark to run 21 miles. If mark runs 2mph faster than terry, how fast d

Alisiya [41]

You need to multiply 21 *2

6 0

3 years ago

A: What are the solutions to the quadratic equation x2+9=0? B: What is the factored form of the quadratic expression x2+9? Selec

Paraphin [41]

Answer:

A. The solutions are $x=3i,\:x=-3i$ .

B. The factored form of the quadratic expression $x^2+9=(x-3i)(x+3i)$

Step-by-step explanation:

A. To find the solutions to the quadratic equation $x^2+9=0$ you must:

$\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}\\\\x^2+9-9=0-9\\\\\mathrm{Simplify}\\\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-9},\:x=-\sqrt{-9}$

$x=\sqrt{-9} = \sqrt{-1}\sqrt{9}=\sqrt{9}i=3i\\\\x=-\sqrt{-9}=-\sqrt{-1}\sqrt{9}=-\sqrt{9}i=-3i$

The solutions are:

$x=3i,\:x=-3i$

B. Two expressions are equivalent to each other if they represent the same value no matter which values we choose for the variables.

To factor $x^2+9$ :

First, multiply the constant in the polynomial by $i^2$ where $i^2$ is equal to -1.

$x^2+9i^2$

Since both terms are perfect squares, factor using the difference of squares formula

$a^2-i^2=(a+i)(a-i)$

$x^2+9=x^2+9i^2=\left(-3i+x\right)\left(3i+x\right)$

5 0

2 years ago