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3 years ago

You deposit $2,000 into a saving account that gains 7% interest each year. How long will it take for your balance to

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

its simple just divide and multipy it by 100%

Step-by-step explanation:

here

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All numbers streets runs parallel to each other both 3rd and 4 the streets are intersected by king Ave

erastova [34]

Answer:

Step-by-step explanation:

King avenue becomes the base

He turns left into the side meet right angle side.

Therefore the hypotenuse cuts through the houses and becomes a measure

Hypotenuse + King Ave B = 3rd Street A

Hypotenuse + 7000mB = 3rd St 5000m A.

H^2= 7m^2B - 5m^2A

√H74 = √25A +√B49

H = √H74 = 8600m

5 0

3 years ago

This is my first time, but my question is what is 7/8 of 20?

Goshia [24]

17.5
Divide 7 by 8 to get .875 and multiply that by 20 to get your answer.

6 0

3 years ago

Read 2 more answers

Some farmers use a circular irrigation method. An irrigation arm acts as the radius of an irrigation circle. How much land is co

SIZIF [17.4K]

Answer:

The land that is covered with an irrigation arm of 370 ft is $430,084\ ft^2$

Step-by-step explanation:

<u>Area of the circle</u>

Being r the radius of a circle, its area can be calculated as follows:

$A=\pi\cdot r^2$

The irrigation arm acts as the radius of the irrigation circle, thus r=370 ft. The land covered is the area under the irrigation arm, thus:

$A=\pi\cdot 370^2$

$A=430,084\ ft^2$

The land that is covered with an irrigation arm of 370 ft is $430,084\ ft^2$

3 0

3 years ago

9/4 divided by 3/8= 9/4 x 8/3

erma4kov [3.2K]

Answer:

9/4:3/8=9/4×8/3=72/12=24/4=6/1=6

7 0

3 years ago

Read 2 more answers

A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet

gayaneshka [121]

Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

$h(x)=10+0.1(x^{1.5})$

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

$h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}$

$s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx$

Let $1+0.0225x=t$

$dx=\frac{1}{0.0225}dt$

$s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt$

$s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}$

$s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}$

$s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)$

s=28.21

Weight of cable= $28.21\times 10.4=293.38$ pound

8 0

3 years ago