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slega [8]
3 years ago
8

Factor x 2 + 29x - 30. (x + 15)(x - 2) (x + 1)(x - 30) (x - 1)(x + 30)

Mathematics
1 answer:
Daniel [21]3 years ago
5 0

Answer:

The correct answer is C) (x - 1)(x + 30)

Step-by-step explanation:

To find this, we need to find factors of the last number (-30) that add up to the middle number (29). In order to do this, list out all of the factors of -30.

-1 and 30

1 and -30

2 and -15

-2 and 15

3 and -10

-3 and 10

Since -1 and 30 satisfy both of the parameters, we put them into their own parenthesis along with x.

(x - 1)(x + 30)

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I need help with #6 please.
Juliette [100K]

9514 1404 393

Answer:

  6. (A, B, C) ≈ (112.4°, 29.5°, 38.0°)

  7. (a, b, C) ≈ (180.5, 238.5, 145°)

Step-by-step explanation:

My "work" is to make use of a triangle solver calculator. The results are attached. Triangle solvers are available for phone or tablet and on web sites. Many graphing calculators have triangle solvers built in.

__

We suppose you're to make use of the Law of Sines and the Law of Cosines, as applicable.

6. When 3 sides are given, the Law of Cosines can be used to find the angles. For example, angle A can be found from ...

  A = arccos((b² +c² -a²)/(2bc))

  A = arccos((8² +10² -15²)/(2·8·10)) = arccos(-61/160) = 112.4°

The other angles can be found by permuting the variables appropriately.

  B = arccos((225 +100 -64)/(2·15·10) = arccos(261/300) ≈ 29.5°

The third angle can be found as the supplement to the other two.

  C = 180° -112.411° -29.541° = 38.048° ≈ 38.0°

The angles (A, B, C) are about (112.4°, 29.5°, 38.0°).

__

7. When insufficient information is given for the Law of Cosines, the Law of Sines can be useful. It tells us side lengths are proportional to the sine of the opposite angle. With two angles, we can find the third, and with any side length, we can then find the other side lengths.

  C = 180° -A -B = 145°

  a = c(sin(A)/sin(C)) = 400·sin(15°)/sin(145°) ≈ 180.49

  b = c(sin(B)/sin(C)) = 400·sin(20°)/sin(145°) ≈ 238.52

The measures (a, b, C) are about (180.5, 238.5, 145°).

7 0
3 years ago
You’re baking a circular cake for a party. Your cake can be any size you want. What will the radius of your cake be and how many
REY [17]
What will the radius of your cake be?

This is a problem of geometry. Given that the cake is circular, the greater the cake the greater the radius of it. So, as shown in the figure 1, the radius will be the distance from the center to any extreme point of the circle.

How many slices will you be able to cut?

The total area of a circle is given by:

A= \pi  r^{2}

We need to fin how many slices will be cut, so let's calculate the area of the circular sector which can be obtained simply applying rule of three, so:

A_{cs}= \pi r^{2}( \frac{\theta}{2\pi})= \frac{r^{2}\theta}{2}

Let's name n the number of slices, if we divide the total area by n this result, each area  must equal, then:

\frac{\pi r^{2}}{n}=\frac{r^{2}\theta}{n}

Finally, we will be able to cut:

n= \frac{2\pi}{\theta} Slices

7 0
4 years ago
Let f(x) = 2x – 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following.
yaroslaw [1]
4. Since f(x) = 2x - 1, we are looking for h(2x - 1), which is (2x-1)^2 + 1. By tha time you expand this you'll get, 4x^2 - 4x + 2. So that is the answer.
5. f(f(x))  = f(2x - 1) = 2(2x - 1) - 1 = 4x - 3
6. h(g(x)) = h(3x) = (3x)^2 + 1 = 9x^2 + 1

8 0
3 years ago
Chen bought a 20-foot chain for $36.50. What is the unit rate of price per foot?
jasenka [17]
About 1.83. You would set up the equation like this
3 0
2 years ago
Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.
alexgriva [62]
Recall that

|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

There are three cases to consider:

(1) When x+2, we have |x+2|=-(x+2) and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4

(2) When x+2\ge0 and x-2, we get |x+2|=x+2 and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)+(x+2)=4

(3) When x-2\ge0, we have |x+2|=x+2 and |x-2|=x-2, so

|x-2|+|x+2|=(x-2)+(x+2)=2x

So

|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x
4 0
3 years ago
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