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3 years ago

How many cubic inches of water can a sphere hold that has a radius of 28in?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer: 18.5 in

Step-by-step explanation:

You need to be more specific

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djyliett [7]

Answer:

do u still need help with this

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3 years ago

Simplificar (x+3) (x-3)

jasenka [17]

(x+3) (x-3)

Multiply the brackets together

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Answer : x^2-9

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3 years ago

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djyliett [7]

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Step-by-step explanation:

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3 years ago

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Mike is saving his emergency funds in an IRA account. Is his strategy a good one?

Leya [2.2K]

No. When he withdraws his funds from the IRA account, there will be taxes applied and he may not be able to get as much as he needs. It is best to keep his emergency funds in a separate account that he can access any time without any extra penalties.

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4 years ago

20 points pls help quick

masha68 [24]

Answer:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

Step-by-step explanation:

We are given the two functions:

$\displaystyle f(x) = 3x^2 + mx +5 \text{ and } g(x) = nx^2 - 4x -2$

And that:

$\displaystyle h(x) = f(x)\cdot g(x)$

With the given conditions that (1, -40) and (-1, 24) satisfy the new function, we want to determine functions f and g.

First, find h:

$\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\ & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}$

Because (1, -40) and (-1, 24) are points on the graph of h, we have that h(-1) = 40 and h(-1) = 24. In other words:

$\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\ -40 &= mn-6m+8n-48 \end{aligned}$

And:

$\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24 & = -mn -2m + 8n +16 \end{aligned}$

Solve the system of equations. Adding the two equations together yield:

$\displaystyle -16 = -8m+16n - 32$

Solve for either m or n:

$\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}$

Substitute this into one of the two equations above and solve:

$\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\ &= (n-2)(n-1) \\ \\ & \end{aligned}$

Therefore:

$\displaystyle n = 2 \text{ or } n = 1$

Solve for m:

$\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or } & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}$

Hence, the values of n and m are either: 2 and 2, respectively; or 1 and 0, respectively.

In conclusion, functions f and g are:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

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2 years ago