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ANEK [815]
3 years ago
5

An art teacher has 24 girls and 40 boys at a soccer camp. The campers will be divided into equal-sized groups. Each group will h

ave the same number of girls and the same number of boys. What is the maximum number of groups that can be formed?
Mathematics
1 answer:
tresset_1 [31]3 years ago
5 0
8 groups is what i suppose
Hope this helps!!! ;-)
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Given f(x)=3x-2 and g(x)=5x, what is f(x)•g(x)
Natasha2012 [34]
F(x)·g(x) = (3x-2)(5x) = 15x^2 -10x
6 0
3 years ago
last week, ezra's sold 1/3 of tangerines and Anthony sold 3/4 of tangerines as ezra's did. how many boxes of tangerines did Anth
Bond [772]

Answer:

¼ box

Step-by-step explanation:

Ezra sold ⅓ box of tangerines.

Anthony sold ¾ as many as Ezra               = ¾ × ⅓ box

Cancel the 3s                                               = ¼ × ¹/₁ box

Multiply numerators and denominators     = ¼ box

Anthony sold ¼ box of tangerines.

6 0
3 years ago
Is y=-½x+2 perpendicular to -2x+1​
BartSMP [9]

Answer:

i think no sorry if im whrong

3 0
3 years ago
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
If you divide one irrational number by another does the result always have to be irrational
Anton [14]
No not always the answer has to be irrational.
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6 0
3 years ago
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