What is the area of the trapezoid below?

Write the expression in repeated multiplication form. Then write the expression as a power. (-8)^3*(-8)^4 The expression in repe

Makovka662 [10]

Given:

The expression is

$(-8)^3\cdot (-8)^4$

To find:

The expression in repeated multiplication form and then write the expression as a power.

Solution:

We have,

$(-8)^3\cdot (-8)^4$

The repeated multiplication form of this expression is

$=[(-8)\cdot (-8)\cdot (-8)]\cdot [(-8)\cdot (-8)\cdot (-8)\cdot (-8)]$

$=(-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)$

Clearly, (-8) is multiplied seven times by itself. So,

$=(-8)^7$

Therefore, the repeated multiplication form of the given expression is $(-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)$ and the expression as single power is $(-8)^7$ .

7 0

3 years ago

Gia is painting her wall. 1 liter of paint can cover 10 square meters. She has a rectangular wall that

garik1379 [7]

5*4 equals 20 so her wall is 20 square meters
Given that one litre can cover 10 square meters, she needs 2 litres of paint to cover the wall

7 0

3 years ago

Helllllp it’s for my lil brother it’s in the pic attached

SSSSS [86.1K]

Answer:

chill out this this you post the 6 times

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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:

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

$b = 17.333\,cm$

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