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3 years ago

5

What’s the surface. area

1 answer:

DENIUS [597]3 years ago

7 0

Answer:

138.61 square centimeter

(Please tell me if I'm wrong)

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A box contain five red balls, a ball is drawn at random, what is the possibility that the ball will be red?

gregori [183]

Answer:

yes because it says it only has red balls it didn't say any other color or something

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3 years ago

Tracy can mow the lawn with a power

rodikova [14]

Answer:

14.29 minutes

Step-by-step explanation:

take the sum of their time 1/20+1/50,then find the reciprocal

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3 years ago

The ratio of the number of cookies to the number of pastries in a box is 3:2. Which table shows the possible amounts of cookies

LenaWriter [7]

The second one is the answer

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4 years ago

Each of my daughters has as many brothers as sisters, but each of my sons has three times as many sisters as brothers. How many

melomori [17]

Put the number of daughters as x and number of sons as y.

For the first half of the statement:
Number of sisters any daughter has = x-1 ( because you can't be your own sibling)
Number of brothers = y

y=x-1.

For the second half:
Number of sisters any son has = x
Number of brothers any son has = y-1

3(y-1)=x

Solve as system of equations.
3y-3=x
3(x-1)-3=x
3x-3-3=x
2x=6
x=3
y=x-1=3-1=2

3 daughters and 2 sons.

Hope this helps!
Your fortune is "You can't have everything...where would you put them all?"

3 0

3 years ago

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:

$y=3$

$r\sin(\theta)=3$

8 0

4 years ago