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solniwko [45]
4 years ago
11

A line passes through the point (-4,-6) and has a slope of 1/2. What is the equation of the line?

Mathematics
1 answer:
strojnjashka [21]4 years ago
7 0

the equation is 1/2x - 6

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How many full 4 2/3 in. Sheets can be cut from 22 1/6 in. Stick
Stolb23 [73]

Answer:

4

Step-by-step explanation:

Because i divided 4 2/3 by 22 1/6 = 4.75 and for full sheets it would be 4

Hope This Helped

5 0
2 years ago
Please help!! I need to know this by today
mario62 [17]
Answer=60

Since the base of the prism is a right triangle, we can use the pythagorean theorem to solve for x.

a²+b²=c²
plug in the data
a²+24²=30²
a²+576=900
subtract 576 from both sides
a²=324
sq root
a=18
So x=18


The formula for the volume of a triangular prism is...

V=(area of the base)*(height)
V=1/2bh*H      where b=base of triangle h=height of triangle, H=height of prism
plug in the data that we know
720=1/2*18*24*H
720=216*H
divide both sides by 216
3 1/3=H

So x=18 and y=3 1/3 or 3.33 repeating

The product of the two

18*3 1/3=60

Answer=60


6 0
3 years ago
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Which fraction is equivalent to 1/2? Use the number line to help answer the question
myrzilka [38]

Answer:

4/8

Step-by-step explanation:

4 0
3 years ago
The larger the number in standard notation, the ____________ the exponent on base 10.
PIT_PIT [208]
Larger I think, sorry if that’s wrong.
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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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