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labwork [276]
3 years ago
11

Simplifying radicals

Mathematics
2 answers:
nevsk [136]3 years ago
6 0
Break the radicand up into a product of known factors to get
--2x^{4} \sqrt[3]{2}
garri49 [273]3 years ago
3 0
2x to the power of 4 times i cube root of 2
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How many cubic centimeters of water can this paper cone cup hold?
weeeeeb [17]

Im probably wrong but 48π cubic centimeters

4 0
3 years ago
Problem number 26 of the Rhind Papyrus says: Find a quantity such that when it is added to StartFraction 1 Over 4 EndFraction of
Rzqust [24]

Answer:

x=12

Step-by-step explanation:

For the given situation , a quantity x is added to \frac{1}{4} x gives 15.

We can set up equation as

x+\frac{1}{4}x=15

Multiply each term by 4 on both sides to get rid the denominator.

It gives,

4 x+x=60

Now, combine like terms

5 x=60

Divide both sides by 5

x=12.

6 0
4 years ago
PLEASE HELP WITH THIS I HAVE NO IDEA HOW TO SOLVE!!!!
VARVARA [1.3K]
Write the vertex form of the equation and find the necessary coefficient to make it work.
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2

You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1

The equation you seek is
.. y = x^2 +6x +7

4 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
4 years ago
The price of a mat is x USD from the start. Write an expression for the price:
Serga [27]

Answer:

Hello good evening friend

5 0
3 years ago
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