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3 years ago

Find the solution(s) to (x + 2)^2 = 49. Check all that apply.

Mathematics

2 answers:

xxMikexx [17]3 years ago

7 0

here's the solution,

=》

$(x + 2) {}^{2} = 49$

=》

$(x + 2) {}^{2} = (7) {}^{2}$

=》

$x + 2 = 7$

=》

$x = 5$

note : value of root 49 = either +7 or -7

so, x = 5 or -9

lianna [129]3 years ago

6 0

Take the square root of both sides of the equation to get (x+2)=7

Then subtract 2 on both sides to get
x=5

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Convert the Cartesian equation x^2 + y^2 = 16 to a polar equation.

RideAnS [48]

Answer:

Problem 1: $r=4$

Problem 2: $r=-2\sin(\theta)$

Problem 3: $r\sin(\theta)=3$

Step-by-step explanation:

Problem 1:

So we are going to use the following to help us:

$x=r \cos(\theta)$

$y=r \sin(\theta)$

$\frac{y}{x}=\tan(\theta)$

So if we make those substitution into the first equation we get:

$x^2+y^2=16$

$(r\cos(\theta))^2+r\sin(\theta))^2=16$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)=16$

Factor the $r^2$ out:

$r^2(\cos^2(\theta)+\sin^2(\theta))=16$

The following is a Pythagorean Identity: $\cos^2(\theta)+\sin^2(\theta)=1$ .

We will apply this identity now:

$r^2=16$

This implies:

$r=4 \text{ or } r=-4$

We don't need both because both of include points with radius 4.

Problem 2:

$x^2+y^2+2y=0$

$(r\cos(\theta))^2+(r\sin(\theta))^2+2(r\sin(\theta))=0$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)+2r\sin(theta)=0$

Factoring out $r^2$ from first two terms:

$r^2(\cos^2(\theta)+\sin^2(\theta))+2r\sin(\theta)=0$

Apply the Pythagorean Identity I mentioned above from problem 1:

$r^2(1)+2r\sin(\theta)=0$

$r^2+2r\sin(\theta)=0$

or if we factor out r:

$r(r+2\sin(\theta))=0$

$r=0 \text{ or } r=-2\sin(\theta)$

r=0 is actually included in the other equation since when theta=0, r=0.

Problem 3:

$y=3$

$r\sin(\theta)=3$

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