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3 years ago

How many times dose 28 go into 6

Mathematics

2 answers:

Rufina [12.5K]3 years ago

6 0

It’s gonna be around 4.6

Ivan3 years ago

3 0

6 goes 4 times into 28 because 6 x 4 = 24

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Which expression is equivalent to 3/2•3/2•3/2•3/2?

serg [7]

The answer is A) 2/3 ^ 4 power

5 0

3 years ago

At a point on the ground 50 feet from the foot of

Likurg_2 [28]

Answer:

$\tan(53) = \frac{x}{50} \\ x = \tan(53) \times 50 \\ x = 66.35 \: feet$

Step-by-step explanation:

I hope that is useful for you :)

5 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:

$0\,\frac{cm^{3}}{s} = \left(10\,\frac{cm^{2}}{s} \right)\cdot \left(\frac{500}{361}\,cm \right)+(361\,cm^{2})\cdot \frac{dh}{dt}$

Then, the rate of change of the height of the box at which is decreasing is:

$\frac{dh}{dt} = -\frac{5000}{130321}\,\frac{cm}{s}$

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

5 0

3 years ago

Solve the equation for the indicated variable.<br> А= 1/2bh for b

Eduardwww [97]

A = 1/2bh

Divide h by both sides.

A/h = 1/2b

Divide 1/2 by both sides.

b = 2a/h

4 0

4 years ago

I need help please, I attached a photo.<br><br>Graph y>1-3x

natali 33 [55]

Your answer is the one on the bottom left

6 0

3 years ago