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2 years ago

Brandy has 8 pounds of candy. She wants to give each of her friends 1/3 pound. To how many friends can Brandy give candy?

Mathematics

1 answer:

marusya05 [52]2 years ago

5 0

He can give 24 friends some candy

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Let f(x)=4x+7 and g(x)=3x-5. find (fog)(-4) A. -9 B. -61 C. -32 D. -17

prohojiy [21]

Answer:

B . (fog)(-4) = - 61

Step-by-step explanation:

Here, given f(x) = 4 x + 7, g(x) = 3 x- 5

Now, by the COMPOSITE FUNCTION:

(f o g) (x) = f (g(x))

Now here: )f o g) ( -4) = f (g(-4))

Simplifying for g(x) = 3 x - 5

g(-4) = 3 (-4) - 5 = -12 - 5 = -17

or, g (-4) = -17

Now, for f(x) = 4 x + 7

f ( g(-4)) = f ( -17) = 4 (-17) + 7

= - 68 + 7 = - 61

or, f ( g(-4)) = - 61

Hence, (fog)(-4) = - 61

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3 years ago

A baby gains about 2 1/5 pounds each month for the first three months after birth.when he was 3 months old,tyler weighed 14 1/10

lyudmila [28]

I believe the answer is 6 9/22.

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3 years ago

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tresset_1 [31]

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3 years ago

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Which of the following are coefficient

malfutka [58]

Answer:

2 and -3

Step-by-step explanation:

because a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).

so in 2x-3y+5, 2 and -3 are the coefficients.

4 0

3 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

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3 years ago