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4 years ago

A(b)(-c), for a=7, b = -2, c= 15

Mathematics

2 answers:

Alexus [3.1K]4 years ago

8 0

A(b)(-c) = 7 (-2)(-15) = 7 (30) = 210. 210 is your answer. Hope this helped!

-Dominant- [34]4 years ago

4 0

You just have to plug in the numbers and solve
7(-2)(-15)
7(30)
210

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3 years ago

Can someone plz answer and explain this?

TiliK225 [7]

Hi,

First let's find the unknown degree in the triangle. We know an entire triangle is 180 degrees so all we have to do is add the sum of the other two interior angles. 55 + 72 = 127. Now let's subtract 127 from 180 to get 53. Now let's solve for x. Since we know the angle beside it is 53, we subtract that from 180 degrees (a line segment is 180 degrees) to get 127.

x = 127 degrees

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4 years ago

The Thornberries are buying a $700,000 home. They have been approved for a 2.51% APR, 25-year mortgage. They made a 25% down pay

fredd [130]

Answer:a. 17,570 b. 48.14

Step-by-step explanation:

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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

A truck is being filled with cube-shaped packages that have side lengths of 1/4 foot. The part of the truck that is being filled

n200080 [17]

Answer:

24000 pieces.

Step-by-step explanation:

Given:

Side lengths of cube = $\frac{1}{4} \ foot$

The part of the truck that is being filled is in the shape of a rectangular prism with dimensions of 8 ft x 6 1/4 ft x 7 1/2 ft.

Question asked:

What is the greatest number of packages that can fit in the truck?

Solution:

First of all we will find volume of cube, then volume of rectangular prism and then simply divide the volume of prism by volume of cube to find the greatest number of packages that can fit in the truck.

$Volume\ of\ cube =a^{3}$

$=\frac{1}{4} \times\frac{1}{4}\times \frac{1}{4} =\frac{1}{64} \ cubic \ foot$

Length = 8 foot, Breadth = $6\frac{1}{4} =\frac{25}{4} \ foot$ , Height = $7\frac{1}{2} =\frac{15}{2} \ foot$

$Volume\ of\ rectangular\ prism =length\times breadth\times height$

$=8\times\frac{25}{4} \times\frac{15}{2} \\=\frac{3000}{8} =375\ cubic\ foot$

The greatest number of packages that can fit in the truck = Volume of prism divided by volume of cube

The greatest number of packages that can fit in the truck = $\frac{375}{\frac{1}{64} } =375\times64=24000\ pieces\ of\ cube$

Thus, the greatest number of packages that can fit in the truck is 24000 pieces.

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3 years ago