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SVETLANKA909090 [29]
2 years ago
13

Ill give brainiest if it was 11:30 am and its now 11:47 am, how many hours was that

Mathematics
1 answer:
Kaylis [27]2 years ago
6 0

Answer:

0 hours and 17 minutes

Step-by-step explanation:

47 - 30 = 17

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What is x divided by x-1 - 1 divided by 2-2x expessed as a single fraction ?
Kay [80]
To simplify into 1 fraction, remember, we must make the denominator the same
x/ (x-1 )  - 1/ (2-2x)
= -x/ -(1-x) - 1/(2-2x)    [see the similarity now? ]  [whatever we multiply for the denominator, as long as we multiply to the numerator, it will be ok]
= -2x / -(2-2x)    + 1/-(2-2x)
= (-2x+1 )/ -(2-2x)
=(2x-1)/(2-2x)
8 0
3 years ago
Ben wrote a check for $52 that was more than his account balance. The bank charges $25 for bounced checks and the local business
MrRa [10]
$97

This is because he will owe the amount of the check $52, plus he will also owe the two fees which will total $45
7 0
3 years ago
Read 2 more answers
What is the range of the function y = x 2?
Leno4ka [110]

Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

  • We know that the range of a function is the set of values of the dependent variable for which a function is defined.

\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

5 0
3 years ago
If h(x)=(fog)(x) and h(x)=³√x+3, find g(x) if f(x) =³√x+2
ipn [44]

Answer:

g(x)=x+1

The problem:

Find g(x) if h(x)=(f \circ g)(x),

h(x)=\sqrt[3]{x+3}, and

f(x)=\sqrt[3]{x+2}.

Step-by-step explanation:

h(x)=(f \circ g)(x)

h(x)=f(g(x))

Replace x in f(x)=\sqrt[3]{x+2} with g(x) since we are asked to find f(g(x)):

\sqrt[3]{x+3}=\sqrt[3]{g(x)+2}

\sqrt[3]{x+1+2}=\sqrt[3]{g(x)+2}

This implies that x+1=g(x)

Let's check:

(f \circ g)(x)

f(g(x))

f(x+1)

\sqrt[3]{(x+1)+2}

\sqrt[3]{x+1+2}

\sqrt[3]{x+3}  which is the required result for h(x).

6 0
3 years ago
Math grade 8<br> Problem 8 show work
Montano1993 [528]

Check the picture below.

4 0
3 years ago
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