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3 years ago

The prime factor of 180

Mathematics

1 answer:

alisha [4.7K]3 years ago

6 0

Answer: 5*2*2*3*3

Step-by-step explanation:

First you divide 180 by 2 and you get 90. Then you proceed to divide that by 2 again and get 45. You will then divide 45 by 3 to get 15. Then divide 15 by 3 again and get 5, which cannot be divided any smaller.

180/2=90

90/2=45

45/3=15

15/3=5

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Which are partial products for 68 × 43?

3241004551 [841]

The correct answer is D? Its the only one that makes sense.

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3 years ago

Find the area of each figure. Round to the nearest tenth if necessary.

Nataly_w [17]

Answer:

220ft^2

Step-by-step explanation:

In red: 12ft - 10ft = 2ft and there 2 sides so 2ft / 2 = 1ft

In blue: 18ft - 12ft = 6ft

The missing area in dark blue: 6ft x 1ft = 6 / 2 = 3

total rectangle area: 12ft x 18ft = 216

216 - (3 + 3) = 216 - 6 = 200 subtract 6, we added that area in

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2 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:

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

$b = 17.333\,cm$

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