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MissTica
3 years ago
7

4(x+x+7)-2x+7-4 by substituting x=1 and x=2

Mathematics
2 answers:
9966 [12]3 years ago
8 0
4(x+x+7)-2x+7-4
Substitute x with 1
4(1+1+7)-2(1)+7-4
=4(2+7)-2+7-4
=4(9)-2+7-4
=36-2+7-4
=34+7-4
=41-7
=34

4(x+x+7)-2x+7-4
Substitute x with 2
4(2+2+7)-2(2)+7-4
=4(4+7)-4+7-4
=4(11)-4+7-4
=44-4+7-4
=40+7-4
=47-4
=43. Hope it help!
yuradex [85]3 years ago
5 0
4(x+x+7) x=1
4(1+1+7)
4+4+28 = 36

(-2x+7-4) x=2
-2(2)+7-4
-4+7-4 = -1

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Taxi A
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2miles £3.50+£3.50=£7.00
Taxi B
2miles £1.25+£4.00=£5.25

Taxi A
3miles £3.50+£5.25=£8.75
Taxi B
3miles £1.25+£6.00=£7.25

Taxi A
4miles £3.50+£7.00=£10.50
Taxi B
4miles £1.25+£8.00=£9.25

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5miles £3.50+£8.75=£12.25
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5miles £1.25+£10.00=£11.25

Taxi A
6miles £3.50+£10.50=£14.00
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Taxi A
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Taxi A
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Taxi B
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^^^
They would have to drive 9 miles for the taxi to cost the same.

Hope this helped, this is the longest way to work it out but also the simplest.
6 0
2 years ago
Mrs. Bhatia's closet consists of two sections, each shaped like a rectangular prism. She plans to buy mothballs to keep the moth
AveGali [126]

Answer:

6.5 boxes

Step-by-step explanation:

Given

See attachment for closet

Required

Determine the number of boxes needed to fill the closet

First, we calculate the volume of the two section.

According to the attachment

The first section has the following dimension:

Width = 5ft

Length = 4ft

Height = 8ft

The second has the following dimension:

Width = 3ft ---- see the last label at the top

Length = 2ft --- This is calculated by subtracting the length of the first section (4ft) from the total length of the closet (6ft) i.e. 6ft - 4ft

Height = 8ft

So: The volume of the closet is:

Volume = W_1 * L_1 * H_1 + W_2 * L_2 * H_2

Volume = 5ft * 4ft * 8ft + 3ft * 2ft * 8ft

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Volume = 208ft^3

The number of box needed is then calculated by dividing the volume of the closet (208ft^3) by the volume of each box (32ft^3)

Boxes = \frac{208ft^3}{32ft^3}

Boxes= \frac{208}{32}

Boxes = 6.5

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2 years ago
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Answer:

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Answer:

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Step-by-step explanation:

Confidence level depicts the probability that the confidence interval actually contains the values of p ( true values of P ) hence

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