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GalinKa [24]
3 years ago
15

There is a box with an area of 112 sq inches and a height of 3.5 inches. What is the width of the box?

Mathematics
1 answer:
Rashid [163]3 years ago
8 0

Answer:

32 inches

Step-by-step explanation:

Because height times width equals area,

we can divide 112 by 3.5 to get the width

So it is 32 inches

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Use the given functions to find f (g(x)), and give the restrictions on x.
torisob [31]

Answer:

f(g(x)) = \frac{10x-9}{3 + 3x}

Step-by-step explanation:

Given

f(x) = \frac{1}{x} - 3

g(x) = \frac{3}{x} + 3

Required

Find f(g(x))

If: f(x) = \frac{1}{x} - 3

Then:

f(g(x)) = \frac{1}{3/x + 3} -3

Solve the denominator (take LCM)

f(g(x)) = \frac{1}{\frac{3+3x}{x}} -3

f(g(x)) = \frac{x}{3 + 3x} - 3

It can be solved further as:

f(g(x)) = \frac{x-3(3 + 3x)}{3 + 3x}

f(g(x)) = \frac{x-9 + 9x}{3 + 3x}

f(g(x)) = \frac{10x-9}{3 + 3x}

4 0
3 years ago
Graph a line with a slope of -3 that contains the point 4,-2
JulsSmile [24]

Answer:

y=-3x+10

Step-by-step explanation:

Use point-slope form first.

y-y1=m(x-x1)

Plug the numbers into that.

y+2=-3(x-4)

Distribute and simplify to convert it to y=mx+b

y+2=-3x+12

y=-3x+10

6 0
3 years ago
I have one more Brainliest
Tresset [83]
Okay, so we need to find the price of the original shirt and the discounted shirt with the 5% tax.

5% (into a real number) is .05

Multiply 20.98 and 15.48 by .05

20.98*.05= 1.049 I am going to round to 1.05

15.48*.05= 0.774 I am going to round to .75

Now we know the prices of both, so we subtract them by each other. 

20.98+1.05= 22.03
15.48+.75= 16.23

22.03-16.23= 5.80

But since we rounded the numbers, I would go with choice D. 

Hope this helps!



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7 0
4 years ago
Plz help me with this question!!!
Alona [7]

Answer:

6.111111111111111111111111111111111111111111111111111 or 6.11 repeating

Step-by-step explanation:

7 0
3 years ago
Solve -5y+8= -3y+10 what is the answer
statuscvo [17]
-3y+10=-5y+8  add 5y to both sides

2y+10=8  subtract 10 from both sides

2y=-2  divide both sides by 2

y=-1
5 0
3 years ago
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