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3 years ago

Find an equivalent percent for 7/10

Mathematics

2 answers:

Yanka [14]3 years ago

5 0

Answer: 70%

Step-by-step explanation: Since a percent is a ratio that compares a number to 100, we first find a fraction that's equivalent to 7/10 that has a 100 in the denominator.

To do this, we set up a proportion.

$\frac{7}{10} = \frac{n}{100}$

Now we can use cross products to find the missing value.

700 = 10n

÷10 ÷10

70 = n

So we can write the fraction 7/10 as 70%.

zubka84 [21]3 years ago

3 0

So, you know that any number with 100 as the denominator will be a fraction. All you have to do is change the 10 you have now to 100. To do that, you have to multiply BOTH numerator and denominator (top and bottom) by 10. That results in 70/100, which is equal to 70%.
The equivalent percent for 7/10 is 70%.

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The solution of the given system of equation is x = -3 and y = 4 respectively.

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A system of linear equations can be defined as a number of equations needed to solve the equations. For n number of variables n number of equations are required.

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the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

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The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

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The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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