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Illusion [34]
3 years ago
11

What is the volume of a sphere that has a radius of 12 mm?

Mathematics
1 answer:
ycow [4]3 years ago
7 0

Answer:

452.16

Step-by-step explanation:

πr^2 ... then substitute the numbers accordingly!

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If you have 12 cups of flour and 3 eggs how many loaves of bread can you make
scoundrel [369]
How much of each is needed to make the bread?
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Find an equation of the line that satisfies the given conditions. through (−5, −7); perpendicular to the line passing through (−
murzikaleks [220]

Answer:

The equation would be y = 2x + 3

Step-by-step explanation:

In order to solve this, we first need to find the slope of the line between (-2, 5) and (2, 3). In order to do this, we use the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (3 - 5)/(2 - -2)

m = -2/4

m = -1/2

Now that we have the original line with a slope of -1/2, we can tell a perpendicular line would have a slope of 2. This is because perpendicular lines have opposite and reciprocal slopes. Now we can use that slope and the given point in point-slope form to get the answer. Be sure to solve for y.

y - y1 = m(x - x1)

y + 7 = 2(x + 5)

y + 7 = 2x + 10

y = 2x + 3

7 0
3 years ago
Lesson 10 Unit 4 ablgebra 1 b test awnsers wth questions
Aleks04 [339]

AND?!?

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Tony makes a 5% commission on all his sales If his commission check is for $152.00 then what were Tony’s sales
shepuryov [24]

Answer:

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Read 2 more answers
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
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