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3 years ago

Rewrite in polar form x^2 + y^2 - 6y - 8 = 0

Mathematics

2 answers:

fredd [130]3 years ago

6 0

the correct answer is B. r^2=6r sin 0+8

Arisa [49]3 years ago

3 0

Answer:

The correct answer option is $r^2-6rsin(\theta)=8$ .

Step-by-step explanation:

We are given the following equation and we are to rewrite it in polar form:

$x^2 + y^2 - 6y - 8 = 0$

$\left \{ {{x^2+y^2=r^2} \atop {y=rsin \theta}} \right.$

Here, we need to make a perfect square trinomial:

$[r^2-6rsin(\theta)+$ ___ $^2]=8$

$6rsin(\theta)$ ---> $2 .r.3.sin \theta$

Completing the square to get:

$r^2-6rsin(\theta)+[3sin(\theta)]^2-[3sin(\theta)]^2=8$

$[r-3sin(\theta)]^2=8+[3sin(\theta)]^2$ --> $[r-3sin(\theta)]^2=8+9sin^2(\theta)$

$r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}$ --> $r=\sqrt{8+9sin^2(\theta)} +3sin(\theta)$

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2 years ago

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2 years ago

Dawn and abed each roll a standard die obtaining a random number from 1 to 6. if the probability that dawns number is larger tha

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Answer:

a+b = 12

Step-by-step explanation:

There are 36 possible outcomes for rolling two dice. Let A be the number rolled by Abed and D be the number rolled by Dawn. The sample space, in the format [A,D], of all of the possibilities in which Dawn's number is larger is:

[5,6]

[4,5] [4,6]

[3.4] [3,5] [3.6]

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7 0

2 years ago

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

Range:

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

3 years ago

Convert to scientific notation: 0.0509.

fomenos

Answer:

5.09 × 10^-2

Step-by-step explanation:

You have to take the number go to the decimal and move it until you get to the first number and in this case, it is 5 put the decimal there. To write it you have to count the number of times you moved the decimal. when doing this it will always be a number times 10^ the number you moved the decimal.

5 0

3 years ago