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3 years ago

5) What point would be the reflection, across the x-axis, for the point (-8,-7)?

Mathematics

1 answer:

melisa1 [442]3 years ago

8 0

Answer: (8,-7)

Step-by-step explanation: Since you are reflecting over the x-axis, the x coordinate stays the same, but the y coordinate does change. Instead of it being negative, the eight would become positive.

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Plz help me with this plz

const2013 [10]

Answer:

Slope: $-\frac{3}{4}$

Y intercept: 7

Equation: $y=-\frac{3}{4}x+7$

Step-by-step explanation:

The slope is found by doing the equation $\frac{rise}{run}$ . When looking at the first two points, it's shown that the rise (difference between the two y values) is -3 (since it decreased by 3) and the run (difference between the two x values) is 4 (since it increased by 4).

This gives us a slope of $-\frac{3}{4}$ .

The y intercept is found by looking at the y value when x = 0. In this picture, it is clear that it is 7.

The equation you want to use is y = mx + b , where b is the y intercept and m is the slope. Just plug in the numbers and you got your equation!

7 0

3 years ago

2. (11.01 MC)

olasank [31]

Answer:

120strip

Step-by-step explanation:

6 0

2 years ago

The graph of the function y=f(x)

jarptica [38.1K]

The range is the high and low points.

So (5, 2]

5 0

2 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$