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Assoli18 [71]
3 years ago
9

Which description matches the graph of the inequality y ≥ |x + 2| – 3? a shaded region above a solid boundary line a shaded regi

on below a solid boundary line a shaded region below a dashed boundary line a shaded region above a dashed boundary line

Mathematics
1 answer:
Alex777 [14]3 years ago
6 0
The given inequality is y ≥ |x + 2| -3.

This inequality may be written two ways:
(a) y ≥ x + 2 - 3
    or
    y ≥ x - 1

(b) y ≥ -x -2 - 3
    or
     y ≥ -x - 5

A graph of the inequality is shown below. The shaded region satisfies the inequality.

Answer: A shaded region above a solid boundary line.

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3 years ago
To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x
Fiesta28 [93]

Given function is

f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

kx^2-5

so RHD= k(-4)^2-5

Now set both equal

k(-4)^2-5=3(-4)+k

16k-5=-12+k

16k-k=-12+5

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k=-\frac{7}{15}

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<u>Hence final answer is -0.47.</u>




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3 years ago
Y=3/2x-3
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Answer:

the answer is (2,-6)

Step-by-step explanation:


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