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4 years ago

Work the following interest problem.

Mathematics

1 answer:

LenKa [72]4 years ago

5 0

R=i/pt
R=625÷13,000
R=0.048*100=4.8%

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What is the equation of the line that passes through the point (-3,-4) and has a slope of 4?

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It is =4x+8 that's the answer

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3 years ago

Find the slope of the line that passes through the points 2,4,6 and 12

kolezko [41]

Answer:

The slope is 2.

Step-by-step explanation:

If you use the formula that is shown in the picture, you can calculate that 12-4/6-2 is the slope. If you simplify, you'll get that 8/4 is the slope which simplifies to 2.

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3 years ago

An eight grader has 18 coins in her pocket. The coins are all dimes and quarters. Their total value is 2.85. Find how much of ea

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3 years ago

Help Due today question Below

statuscvo [17]

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3 years ago

Read 2 more answers

A box witxh a height (x+5) cm has a square base with side x com. A second box with height (x+2) com has a square base with side

Eva8 [605]

Answer:

x=5.37 cm

Step-by-step explanation:

we know that

The volume of the box is

$V=Bh$

where

B is the area of the base of the box

h is the height of the box

we have that

The area of the base is

$B=x^2\ cm^2$

$h=(x+5)\ cm$

The volume of the Box 1

is equal to

$V_1=x^2(x+5)\ cm^3$

$V_1=(x^3+5x^2)\ cm^3$

we have that

The area of the base is

$B=(x+1)^2\ cm^2$

$h=(x+2)\ cm$

The volume of the Box 2

is equal to

$V_2=[(x+1)^2(x+2)]\ cm^3$

$V_2=[(x^2+2x+1)(x+2)]\ cm^3$

$V_2=(x^3+2x^2+x+2x^2+4x+2)\ cm^3$

$V_2=(x^3+4x^2+5x+2)\ cm^3$

Equate the equation of Volume 1 to the equation of Volume 2

$(x^3+5x^2)=(x^3+4x^2+5x+2)$

$(5x^2)=(4x^2+5x+2)$

$5x^2-4x^2-5x-2=0$

$x^2-5x-2=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is x=5.37 cm

see the attached figure

8 0

4 years ago