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Setler79 [48]
3 years ago
14

The inequality x < 9 or x ≥ 14 can be used to represent the hourly wage, x, of each employee at a store. Which are possible v

alues for x? Select two options.
$8
$9
$11
$13
$14
Mathematics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

$11, $13, $14

Step-by-step explanation:

all of those answers are greater than $9 but less than or equal to $14.

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Write the slope intercept form of the equation of the line through the given points and slope 1,1 and slope of -3/5
Maslowich

Slope is already given in the question so we only need to solve for the y-intercept.

Slope intercept form: y = mx + b

m = slope

b = y-intercept

-3/5 = -0.6

1 = -0.6(1) + b

1 = -0.6 + b

1 + 0.6 = -0.6 + b + 0.6

1.6 = b

Now, write in slope-intercept form.

y = -3/5x - 1.6

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Best of Luck,

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7 0
3 years ago
I need help with this question
prisoha [69]

Answer:

100m^2

Step-by-step explanation:

square has equal sides.

To find the area of the square, the formula is:

Area= length times width

So for square ABCD, 10*10=100m^2

8 0
3 years ago
Please help! I attached the question below.
kompoz [17]

Answer:

\frac{2(c+2)}{c(c-2)}

Step-by-step explanation:

\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }

Identity used:

a^{2}-b^{2}=(a-b)(a+b)

\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }

Now let us divide the modified expressions:

\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})} ÷ \frac{(c-2)^2}{2(6c^{3}+15c^{2}) }

we get:

\frac{2(c+2)}{c(c-2)}

5 0
3 years ago
I need clean and clear answers please!<br><br> Thank you
lina2011 [118]
Okay so we get: subtract m + 8 from 5m + 11

5m + 11 = 16 + m or 16m

16m - 8 +m or 8m = 8m or 8 + m

5 0
3 years ago
Question 12 Multiple Choice Worth 1 points)(07.05 MC)1086Which of the following functions best represents the graph?fix)=(x-2)(x
vlada-n [284]

The graph of a function is given. It is required to find the function that represents the graph.

Notice from the graph that the function has roots or zeros at -3,-2, and 2.

Hence, the function takes the form:

\begin{gathered} f(x)=a(x-(-3))(x-(-2))(x-2) \\ \Rightarrow f(x)=a(x+3)(x+2)(x-2) \end{gathered}

Where a is a constant to be determined.

Notice that the given graph passes through the point (0,-12).

Substitute x=0 and f(x)=-12 into the equation of the function to find a:

\begin{gathered} -12=a(0+3)(0+2)(0-2) \\ \Rightarrow-12=a(3)(2)(-2) \\ \Rightarrow-12=-12a \\ \text{ Swap the sides of the equation:} \\ \Rightarrow-12a=-12 \\ \text{ Divide both sides by }-12: \\ \Rightarrow\frac{-12a}{-12}=\frac{-12}{-12} \\ \Rightarrow a=1 \end{gathered}

Hence, the function is:

\begin{gathered} f(x)=1(x+3)(x+2)(x-2) \\ \Rightarrow f(x)=(x+2)(x+3)(x-2) \end{gathered}<h2>The third option is the answer: f(x)=(x+2)(x+3)(x-2).</h2>
5 0
1 year ago
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