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3 years ago

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What is the quotient StartFraction 15 p Superscript negative 4 Baseline q Superscript negative 6 Baseline Over negative 20 p Sup

erscript negative 12 Baseline q Superscript negative 3 Baseline EndFraction in simplified form? Assume p not-equals 0, q not-equals 0.

1 answer:

Kitty [74]3 years ago

8 0

Answer:

$-\frac{3p^8}{4q^3}$

Step-by-step explanation:

$\frac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}$ can be broken down into three fractions, coefficients, powers of p, powers of q.

$-\frac{15}{20}\cdot\frac{p^{-4}}{p^{-12}}\cdot\frac{q^{-6}}{q^{-3}}$

Simplify the first fraction, then simplify the others by subtracting numerator exponents minus denominator exponents.

$-\frac{3}{4} p^{-4-(-12)} q^{-6-(-3)} =-\frac{3}{4} p^8 q^{-3} = \alpha -\frac{3p^8}{4q^3}$

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For the functions f and g, find (f o g)(8). f(x) = 17x^2 - 10x, g(x) = 9x - 9

Andreyy89

Hey mate. Here is your answer.

Set up the composite function and evaluate.

g (17x^2 - 10x) = 153x^2 - 90x - 9

Hope this helps.

5 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

Check the consistency of 2x-3y=7 and x-2y+1=0

Furkat [3]

Answer:

x = 17, y = 9

or as an ordered pair, (17, 9).

The equations are consistent.

Step-by-step explanation:

2x - 3y = 7

x - 2y = -1 Multiply this equation by -2:

-2x + 4y = 2 Add this to the first equation:

y = 9.

Plug y = 9 into the second equation:

x - 2*9 = -1

x = -1 + 18 = 17.

3 0

3 years ago

Stuck on question 1, can anyone help???

poizon [28]

'A' is the square root of 25. That's 5, so take A=5 with you
as you go to the next step.

B is A³. A³ means (A x A x A). We know that 'A' is 5, so 'B' is (5x 5 x 5) = 125 .
Take B=125 with you to the next step.

'C' is B - 25. We know that 'B' is 125. So C = (125 - 25) = 100 .
Take C=100 with you to the next step.

'D' is the square root of 'C'. We know that C=100, so D = √100 .
The square root of 100 is 10, so D=10.
Take D=10 with you to the next step .

'E' is D+39. We know that D=10. So E=(10+39) = 49 .
Take E=49 with you to the last step.

'F' is the square root of 'E'. We know that E=49.

7 0

3 years ago

Read 2 more answers

Solve for a in terms of F and m: F=ma

Elena-2011 [213]

To solve for a, you must get a by itself on one side of the equation. Now, a is "joined" to m by multiplication. To "separate" them, use the inverse operation, division.

Divide both sides of the equation by m.

$F=ma \\ \frac{F}{m}=\frac{ma}{m} \\ \frac{F}{m}=a \\ a=\frac{F}{m}$

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3 years ago