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katrin [286]
3 years ago
14

I need help. Someone help please. (photo)​

Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

Step-by-step explanation:

In order to solve all of these problems, you have to first create an equation. The typical format for an equation looks like this: y = mx + b

m is the slope and b is the y-intercept.

In order to create an equation, you have to find the slope and y-intercept of the function. You can find the slope using this expression: \frac{y2-y1}{x2-x1} To find the y-intercept, you use a known point and the slope in the equation to solve for b.

Now that we know this, we can solve the problems.

1a.

To create the equation, let's find the slope using points (1, 15) and (2, 30)

\frac{30 - 15}{2 - 1} = 15/1 = 15

The equation now looks like this: h = 15w + b

Let's use the point (1, 15) from the table to find b

15 = 15(1) + b

15 = 15 + b

subtract 15 from both sides the isolate the b

0 = b

There is no y-intercept for this equation, so the equation looks like this:

h = 15w

1b. We know the equation is h = 15w, so all we have to do is input the 9 into the equation.

h = 15(9)

h = 135

2a. We'll use the points (1, 8) and (2, 16) to find the slope

\frac{16 - 8}{2-1} = 8/1 = 8

The slope is 8.

Since it's impossible to sell any houses in 0 months, we already know that there is no y-intercept, so the equation looks like this:

h = 8m

2b. Put the value of 15 inside the equation and solve.

h = 8(15)

h = 120

3a. We'll use the points (1, 35) and (2, 70) to find the slope.

\frac{70 - 35}{2 - 1} = 35/1 = 35

The slope is 35.

It's impossible to see any movies in 0 months, so there is no y-intercept. The equation looks like this:

c = 35m

3b. We know that there are 12 months in a year, so put 12 into the equation and solve.

c = 35(12)

c = 420

4a. We'll use the points (1, 8) and (2, 16) to find the slope

\frac{16 - 8}{2 - 1} = 8/1 = 8

The slope is 8.

It's impossible for there to be a cost if there are no hats, so the y-intercept must be 0. The equation looks like this:

c = 8h

4b. Put the 30 hats into the equation and solve.

c = 8(30)

c = 240

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3 0
3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\


\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
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\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\
3x^3-5x^2-14x-4=0\\
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~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
2 years ago
200-[(50-4)×3+5]<br><img src="https://tex.z-dn.net/?f=200%20-%20%28%2850%20-%204%29%20%5Ctimes%203%20%2B%205%29" id="TexFormula1
lbvjy [14]
200 - [ ( 50 - 4 ) × 3 + 5 ]
= 200 - [ 46 × 3 + 5 ]
<Put brackets around the 2 number that enclose the times sign as a reminder to do that first>
= 200 - [ ( 46 × 3 ) + 5 ]
= 200 - [ 138 + 5 ]
= 200 - (143)
= 57

Hope this helps!
3 0
3 years ago
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