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Feliz [49]
2 years ago
8

Given f(x) = √x and g(x) = x+2 find the domain of f(g(x)).

Mathematics
1 answer:
REY [17]2 years ago
8 0

Answer:

E) None of the above

Step-by-step explanation:

Given f(x)=\sqrt{x} and g(x)=x+2, then f(g(x))=\sqrt{x+2}. This shifts the graph of the parent function f(x)=\sqrt{x} 2 units to the left. Thus, the domain of the function is [-2,\infty), making none of the above true.

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What is the domain of the function.....<br> {TIMED PLS HURRY}
timofeeve [1]

try c im sorry if im wrong i tried

6 0
2 years ago
Matt gets paid an hourly rate of $12.50 for the first 40 hours he works in a week. After 40 hours , Matt gets paid at an overtim
lukranit [14]

12.50 x 40 = 500; this is his first week pay from $12.50 an hour for 40 hours  

(15 x 55) + 100 = 925; this is the second pay from $15 an hour for 55 hours and an additional $100


Hope this helps! (If not correct me if its wrong xd)

5 0
2 years ago
Rewrite the function y= x^2 + 14x + 4 in vertex form
Step2247 [10]
Vertex form is basically commplete the square
y=a(x-h)^2+k


y=x^2+14x+4
take 1/2 of 14 and square it, (7^2=49)
add that and its negative to right side
y=x^2+14x+49-49+4
factor perfect squaer
y=(x+7)^2-49+4
y=(x+7)^2-45

answer is A
4 0
3 years ago
Given H(6,7) and I (-7,-6) if point G lies 1/2 of the way long line segment HI, Santiago argues that point G is located at the o
amid [387]
A point that is half way of line segment is the midpoint

Let (6, 7) be (x₁, y₁) and (-7, -6) be (x₂, y₂)

Mid point of HI = (\frac{x_1+x_2}{2},  \frac{y_1+y_2}{2})
Mid point of HI = (\frac{6+-7}{2},  \frac{7+-6}{0})
Mid point of HI = (- \frac{1}{2} ,  \frac{1}{2})

Santiago's statement is not correct
4 0
2 years ago
Which of the following is an equivalent equation obtained by completing the square of the expression below? x^2+6x−8=0 A (x+3)^2
valentina_108 [34]

Answer:

Solving the equation x^2+6x-8=0 by completing the square method we get \mathbf{(x+3)^2=17}

Option B is correct option.

Step-by-step explanation:

We need to solve the equation x^2+6x-8=0 by completing the square method.

For completing the square method: we need to follow: a^2+2ab+b^2=(a+b)^2

We are given:

x^2+6x-8=0

Solving by completing the square

x^2+2(x)(?)+(?)^2-8=0

We need to find ? in our case ? is 3 because 2*3= 6 and our middle term is 6x i,e 2(x)(3)=6x.

So, adding and subtracting (3)^2

x^2+2(x)(3)+(3)^2-8-(3)^2=0\\(x+3)^2-8-9=0\\(x+3)^2-17=0\\(x+3)^2=17

So, solving the equation x^2+6x-8=0 by completing the square method we get \mathbf{(x+3)^2=17}

Option B is correct option.

7 0
2 years ago
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