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3 years ago

Convert the following percent to a decimal and then to a common fraction. 87.5%

Mathematics

2 answers:

nika2105 [10]3 years ago

8 0

Convert to a fraction by placing the decimal number over a power of 10
so your answer would be 7/8

andriy [413]3 years ago

3 0

Your answer would be .875 and 87.5/100

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Find the square root of 13²

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The answer to that would be “169”

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4 years ago

A line with a slope of 5 passes through the point (2, 5). What is its equation in

Maru [420]

Answer: y = 5x - 5

Step-by-step explanation:

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3 years ago

Read 2 more answers

Tammy borrowed $5000 at a rate of 12% compounded monthly. Assuming she makes no payments, how much will she owe after 9 years?

Dominik [7]

Answer:

$14,644.63

Step-by-step explanation:

To solve this problem we can use the compound interest formula which is shown below:

$A=P(1+\frac{r}{n} )^{nt}$

P = initial balance

r = interest rate

n = number of times compounded annually

t = time

First change 12% into a decimal:

12% -> $\frac{12}{100}$ -> 0.12

Lets plug in the values:

$A=5,000(1+\frac{0.12}{12})^{9(12)}$

$A=14,644.63$

Tammy will own $14,644.63 after 8 years,

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3 years ago

What is the root of the quadratic equation y= (x-4) (x+7)

pishuonlain [190]

Root are also known as what x is equal to. So, to find x you would need to set (x-4)=0 and (x+7)=0
X=4. X=-7

4 0

3 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

3 years ago