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Dovator [93]
3 years ago
8

How do reference angles work and how do you find them?

Mathematics
1 answer:
andre [41]3 years ago
8 0

A 200-degree angle is between 180 and 270 degrees, so the terminal side is in QIII.

Do the operation indicated for that quadrant. Subtract 180 degrees from the angle, which is 200 degrees.

You find that 200 – 180 = 20, so the reference angle is 20 degrees.

hope this helped!

o(* ̄▽ ̄*)ブ

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How to solve 3G+6(-g+(-5))
9966 [12]
First of all get rid of the parentheses,so it will be 3G-6g-5=-3g-5 I hope that I answered correctly
5 0
3 years ago
(X^2+x-6)(2x^2+4x) factor completely
MissTica

Answer:

2x\left(x-2\right)\left(x+3\right)\left(x+2\right)

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring x^2+x-6.

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For ax^2+bx+c we need to find u, v > u\cdot \:v=a\cdot \:c and u+v=b which we will group into \left(ax^2+ux\right)+\left(vx+c\right).

The values we have are...a=1,\:b=1,\:c=-6 and u\cdot v=-6,\:u+v=1.

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow  \mathrm{True}

\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow  \mathrm{False}

Therefore u=3,\:v=-2.

Now we want to \mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right). Which is \left(x^2-2x\right)+\left(3x-6\right).

Now we must factor x from x^2-2x.

Lets apply the exponent rule of a^{b+c}=a^ba^c which means x^2=xx.

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of 3x-6.

Rewrite 6 as 3 * 2. 3x-3\cdot \:2.

Now factor out the common term of 3 > 3\left(x-2\right).

x\left(x-2\right)+3\left(x-2\right) > Factor out the common term of x - 2 > \left(x-2\right)\left(x+3\right).

Now lets factor 2x^2+4x.

Apply the previous exponent rule of a^{b+c}=a^ba^c > x^2=xx > 2xx+4x.

Now rewrite 4 as 2 * 2 > 2xx+2\cdot \:2x.

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get 2x\left(x-2\right)\left(x+3\right)\left(x+2\right).

Hope this helps!

7 0
3 years ago
What are the domain and range of the function f(x)=-3(x-5)2 +4?
Xelga [282]

Answers:

Domain is (-\infty, \infty)

Range is (-\infty, 4]

============================================

Explanation:

We can replace x with any real number we want. We don't have any restrictions to worry about since there are no division by zero issues for instance. Also, there isn't any issues of things like taking the square root of a negative number.

Therefore, the domain is the set of all real numbers which translates to the interval notation of (-\infty, \infty)

This interval notation can be thought of as -\infty < x < \infty

----------------------------

The range on the other hand isn't the set of all real numbers. It might help to graph this parabola (see below). You should see that the highest point occurs at the vertex (5, 4). This then tells us that the largest y can get is y = 4.

In other words, y = 4 or y can be smaller than this.

In symbols, we would say the range is y \le 4 and that translates to the interval notation of (-\infty, 4]

The curved parenthesis always goes with either infinity. The square bracket says "include the endpoint 4 as part of the interval".

4 0
3 years ago
Which expression has the same value as -18-(-9)?<br> 0 - 18+2<br> -12-(-3)<br> - 1945<br> O-8-(-4)
olganol [36]

Answer:

the answer is -12-(-3)

Step-by-step explanation:

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Sasha is selling t-shirts. She uses the function f(x) = 2x + 8 to determine her sales, where
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Answer:

f(x)=56

Step-by-step explanation:

2(24)+8=56

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3 years ago
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