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pickupchik [31]
3 years ago
9

Please show me how to COMPLETELY factor this expression.

Mathematics
1 answer:
ss7ja [257]3 years ago
6 0
The answer is
(z+14)(z-3)
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What are all the possible rational roots of f(0) = 2x^4+ x^3+ 4?
ElenaW [278]

Answer: the answer is 4, just 4

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3 years ago
What expression is equivalent to 6+3(n-4)-8+2n
lorasvet [3.4K]

Answer:

Step-by-step explanation:

Comment

Begin by removing the brackets.

6+3(n-4)-8+2n

6 + 3n - 12 - 8 + 2n                 Collect like terms

3n + 2n + 6 - 12 - 8                 Combine

5n - 14

Answer

5n - 14

4 0
2 years ago
At 6:00 p.m. the temperature was –5°F outside. By midnight the temperature had dropped 12 degrees. What was the temperature at m
USPshnik [31]
-5 - 12 = -17

Starts at -5 degrees, then it drops another 12, so you subtract 12 from -5.
6 0
3 years ago
Read 2 more answers
How do you solve this equation . Q=3a+5ac
mr_godi [17]
A "solution" would be a set of three numbers ... for Q, a, and c ... that
would make the equation a true statement.

If you only have one equation, then there are an infinite number of triplets
that could do it.  For example, with the single equation in this question,
(Q, a, c) could be (13, 1, 2) and they could also be  (16, 2, 1).
There are infinite possibilities with one equation.

In order to have a unique solution ... three definite numbers for Q, a, and c ...
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8 0
3 years ago
an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
3 years ago
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