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lbvjy [14]
4 years ago
12

Find the quantity represented by each percent. 120% of $590

Mathematics
2 answers:
Ivanshal [37]4 years ago
6 0
120% is same as 120/100
120/100*590=708
answer is 708
Stella [2.4K]4 years ago
5 0
120% of 590=1.20*590=708.
For x% of a number y, the answer is x/100*y.
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If , n^-3 = 1/8 then n would be which of the following? -1/2 -2 2 1/2
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2

Step-by-step explanation:

{n}^{ - 3}  =  \frac{1}{8}  \\  \\  {n}^{ - 3}  =  \frac{1}{ {2}^{3} }  \\  \\  {n}^{ - 3}  = {2}^{ - 3}.. (\because \frac{1}{a^n} = a^{-n}) \\  \\ n = 2 \\(\because exponents \: are \: same\: hence\: \\bases\: will \: also\: be\: equal)

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Which ratio expresses the scale used to create this drawing?<br> 1 square=10 yards
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option B

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4 0
3 years ago
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
nekit [7.7K]

Answer:

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}

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

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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3 years ago
What is 7/10 in a decimal and percent
Ksivusya [100]
In a decimal it is 0.70 and in a percent it is 70%
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3 years ago
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