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topjm [15]
3 years ago
13

PLEASE HELP!!! WHICH OF THE FOLLOWING FUNCTIONS IS GRAPHED BELOW?

Mathematics
1 answer:
SashulF [63]3 years ago
7 0

Answer: D.

Step-by-step explanation:

For an absolute value function, the vertex of f(x) = |x+h| + k is defined as the point (-h, k) for the coordinate (x, y).

When x is equal to negative h, the value for x and value for h effectively cancel out, and only the positive k remains, hence the vertex being (-h, k).

The function given has a vertex at (2, 3). We know that the vertex of an absolute function is (-h, k), so h must equal -2 and k must equal 3.

The equation:

f(x) = |x-2| + 3

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Step-by-step explanation:

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Read 2 more answers
Identify the type of hypothesis test below. H0:X=10.2, Ha:X&gt;10.2 Select the correct answer below: The hypothesis test is two-
Ganezh [65]

Answer:

The hypothesis test is right-tailed

Step-by-step explanation:

To identify a one tailed test, the claim in the case study tests for the either of the two options of greater or less than the mean value in the null hypothesis.

While for a two tailed test, the claim always test for both options: greater and less than the mean value.

Thus given this: H0:X=10.2, Ha:X>10.2, there is only the option of > in the alternative claim thus it is a one tailed hypothesis test and right tailed.

A test with the greater than option is right tailed while that with the less than option is left tailed.

4 0
3 years ago
Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
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