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Mars2501 [29]
3 years ago
12

After three plays in which a football team lost 7 yd, gained 3 yd, and

Mathematics
1 answer:
True [87]3 years ago
4 0

Answer:

25 Yard line was the previous location.

Step-by-step explanation:

-7+3-1= -8+3= -5

30-5=25.

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#82 will give brainliest to best answer!​
Fofino [41]

Answer:

\frac{6}{k-6}

Step-by-step explanation:

First, we can factor all of the following equations to turn that weird, huge looking thing into \frac{(k+6)(k-6)}{(k-6)(k-10)} ÷ \frac{(k-6)^2}{k(k-6)} × \frac{6(k-10)}{k(k + 6)}. We know that division is simply multiplication by the reciprocal, so that whole equation will turn into \frac{(k+6)(k-6)}{(k-6)(k-10)} × \frac{k(k-6)}{(k-6)^2} × \frac{6(k-10)}{k(k+6)}. Now we can cancel out some values if they are both in the numerator and denominator, which will turn that still huge looking thing into \frac{6}{k-6} which is our final answer, as it cannot be simplified further.

Hope this helped! :)

7 0
2 years ago
Read 2 more answers
Simplify: (7x4 – 7x3 + 7x2 + 2x) + (-5x4 + 7x3 + 5x2 – 9)​
ikadub [295]

Answer:

Step-by-step explanation:

8 0
2 years ago
On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3)
goldfiish [28.3K]

Answer:

f(x) > 0 over the interval (-\infty,-4)

Step-by-step explanation:

If  f(x)  is a continuous function,  and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.

Then the function was above the x axis (larger than zero) from - \infty, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.

There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from - \infty to -3, since between the  values -4 and -3 the function adopts negative values.

6 0
3 years ago
Read 2 more answers
I need help understanding square roots
Vlad [161]

I wonder if you mean to write \sqrt[3]{256} in place of 3\sqrt{256}...

If you meant what you wrote, then we have

\sqrt{256}=\sqrt{16^2}=16

3\sqrt{256}=3\cdot16=48

If you meant to write \sqrt[3]{256} (the cube root of 256), then we could go on to have

\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4

6 0
3 years ago
Please post an provide Simplify 5p^2\5p^3
Rom4ik [11]
( p - 3) or (2p + 1)
5 0
2 years ago
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