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4 years ago

Use only an associative property to rewrite the following expression. (Pk)•r=

Mathematics

1 answer:

MrMuchimi4 years ago

5 0

(Rk)•p would be the answer
All you need to do is rearrange the variables

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50 POINTS I WILL MARK BRAINLIEST PLZ HELP ITS TIMED

oksano4ka [1.4K]

Answer:

The given polygon has 4 vertices and, 4 non-parallel and non equal sides. So the given polygon is a quadrilateral.

Step-by-step explanation:

The given vertices of are (–2, 4), (5, 4), (3, –3), and (–2, 0).

i found this on web

6 0

3 years ago

Read 2 more answers

What number has a 9 with a value ten times as many as the 9 in 39,154

Tomtit [17]

Answer:

Any number with 9 in the ten-thousands place. 90,000 is one such number.

Step-by-step explanation:

The 9 in 39,154 is in the thousands place. Its value is 9,000. In order for the 9 in a number to have a value 10 times that, or 90,000, the 9 must be in the ten-thousands place.

There are an infinite number of such numbers. We suspect you have a list you are to choose from. Pick the number with 9 where it is in the number 90,000.

8 0

3 years ago

The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 10

serious [3.7K]

Answer:

The rate of change of the height of the box at which is decreasing is $\frac{5000}{130321}$ centimeters per second.

Step-by-step explanation:

From Geometry the volume of a rectangular box ( $V$ ), measured in cubic centimeters, with a square base is modelled by the following formula:

$V = A_{b}\cdot h$ (Eq. 1)

Where:

$A_{b}$ - Area of the base, measured in square centimeters.

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

The height of the box is cleared within the formula:

$h = \frac{V}{A_{b}}$

If we know that $V = 500\,cm^{3}$ and $A_{b} = 361\,cm^{2}$ , then the current height of the box is:

$h = \frac{500\,cm^{3}}{361\,cm^{2}}$

$h = \frac{500}{361}\,cm$

The rate of change of volume in time ( $\frac{dV}{dt}$ ), measured in cubic centimeters per second, is derived from (Eq. 1):

$\frac{dV}{dt} = \frac{dA_{b}}{dt}\cdot h + A_{b}\cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA_{b}}{dt}$ - Rate of change of the area of the base in time, measured in square centimeters per second.

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

If we get that $\frac{dV}{dt} = 0\,\frac{cm^{3}}{s}$ , $\frac{dA_{s}}{dt} = 10\,\frac{cm^{2}}{s}$ , $h = \frac{500}{361}\,cm$ and $A_{b} = 361\,cm^{2}$ , then the equation above is reduced into this form: