The square below represents one whole.<br> What percent is represented by the shaded area?

Given the function ƒ(x) = 3x + 5, find ƒ(4) and x such that ƒ(x) = 38.

Sonja [21]

Answer:

just plug 4 inside f(x)=3(4)+5

3x+5=38

3x=33

Divid by 3 on both sides

x=11

Step-by-step explanation:

3 0

3 years ago

Simplify the expression 8h - 57-3m by adding or subtracting the like term

tigry1 [53]

Answer:

You cannot simplify this expression.

Step-by-step explanation:

There are no like terms, meaning this is as simplified as it can get

4 0

3 years ago

Read 2 more answers

Help pleaseeeeeeee!!!

Kamila [148]

Answer:

B or the second option

Step-by-step explanation:

the rules of math state that you must not have a square root in the denominator so in order to get rid of it multiply the top and bottom by the square root on the bottom. Any square root times itself takes away the square root symbol

6 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: