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3 years ago

Lol suisitctftf ysyajwnw yakahad gwhnqga

Mathematics

1 answer:

Lynna [10]3 years ago

4 0

Im not sure what this is, or how to answer it

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Please answer this question. I’m giving brainliest. Both of them, please :)

Solnce55 [7]

Answer: where's the question

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the range of the function f(x) = |x| +2?

kykrilka [37]

Imagine the graph of f(x)=|x|. This can only output positive values ( $x \geq 0$ ).

Then f(x)=|x|+2 is the same graph shifted upwards by two units meaning it can only output values bigger or equal to than 2. So the range is $x \geq 2$

5 0

3 years ago

Using these coordinates, find the slope. (0, -6), (3, -2)

seropon [69]

Answer:

The slope between the points (0, -6) and (3, -2) will be:

$m=\frac{4}{3}$

Step-by-step explanation:

Given the points

(0, -6)

(3, -2)

The slope between the points (0, -6) and (3, -2) will be:

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(0,\:-6\right),\:\left(x_2,\:y_2\right)=\left(3,\:-2\right)$

$m=\frac{-2-\left(-6\right)}{3-0}$

$m=\frac{4}{3}$

Therefore, the slope between the points (0, -6) and (3, -2) will be:

$m=\frac{4}{3}$

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3 years ago

Determine the zeros of the expression -16t^2-8t+24.

Lina20 [59]

$-16t^2-8t+24=-8(2t^2+t-3)=-8(2t^2+3t-2t-3)\\\\=-8[t(2t+3)-1(2t+3)]=-8(2t+3)(t-1)\\\\\text{the zeros}\\\\2t+3=0\to t=-1.5\\t-1=0\to t=1\\\\Answer:\ \boxed{-1.5\ and\ 1}$

6 0

3 years ago

Help me out please!!!!

DedPeter [7]

Answer:

Answer (e)

Step-by-step explanation:

Note that this graph has a zero at x = 1 and a vertical asymptote at x = -4. Only Answer (e) corresponds to that. Try letting x = 1. y is zero, correct? Try letting x = -4. y is undefined, correct?

7 0

3 years ago