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2 years ago

Find the range of the graphed function.

Mathematics

1 answer:

sergiy2304 [10]2 years ago

3 0

Answer:

The answer is C

Step-by-step explanation:

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Explain how you can find the area of a polygon

anyanavicka [17]

Answer:

put a ruler in it and there it is

7 0

3 years ago

What is the 8th term of 6,2,2/3

devlian [24]

Answer:

2/729

Step-by-step explanation:

I think that is the answer I hope it helps

3 0

3 years ago

The family of functions y = ce^{-2x} + e^{-x} is solution of the equation y' + 2y = e^{-x}. Find the constant c which defines th

Rufina [12.5K]

Answer:

$c = 1190.2$

Step-by-step explanation:

We have the following solution:

$y(x) = ce^{-2x} + e^{-x}$

We want to find the value of x that satisfies the following condition:

y(3) = 3.

This means that when x = 3, y = 3. So

$y(x) = ce^{-2x} + e^{-x}$

$y(3) = ce^{-2*3} + e^{-3}$

$3 = ce^{-6} + e^{-3}$

$3 = ce^{-6} + 0.0498$

$ce^{-6} = 2.9502$

$c = \frac{2.9502}{e^{-6}}$

$c = 1190.2$

4 0

3 years ago

2 2/3 + 12 6/8 how do I work this

barxatty [35]

Answer:

$\large\boxed{2\dfrac{2}{3}+12\dfrac{6}{8}=15\dfrac{5}{12}}$

Step-by-step explanation:

$2\dfrac{2}{3}+12\dfrac{6}{8}=2\dfrac{2}{3}+12\dfrac{6:2}{8:2}=2\dfrac{2}{3}+12\dfrac{3}{4}\\\\\text{Find}\ LCD:\\\\\text{List of multiples of 3:}\ 0,\ 3,\ 6,\ 9,\ \boxed{12},\ 15,\ ...\\\text{List of multiples of 4:}\ 0,\ 4,\ 8,\ \boxed{12},\ 16,\ ...\\\\12=3\cdot4\\\\\text{therefore}\\\\\dfrac{2}{3}=\dfrac{2\cdot4}{3\cdot4}=\dfrac{8}{12}\\\\\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\\\\2\dfrac{2}{3}+12\dfrac{3}{4}=2\dfrac{8}{12}+12\dfrac{9}{12}=(2+12)+\dfrac{8+9}{12}=14+\dfrac{17}{12}\\\\=14+1\dfrac{5}{12}=15\dfrac{5}{12}$

3 0

3 years ago

The numbers on a football field indicate 10 yard increments. You walk around the perimeter of a football field between the pylon

Roman55 [17]

Answer:

5,333 square yards.

Step-by-step explanation:

Football fields have standard 100 yards length, if you walk the perimeter of a football field you will walk the twice the length and twice the width. Therefore, the width of the field can be found as follows:

$306.667 = 2L +2W\\W = \frac{306.667 - 2*100}{2} \\W= 53.33$

The area of the playing field is the product of the length by the width

$A = L*W\\A= 100*53.33\\A=5333$

The area of the playing field is 5,333 square yards.

5 0

3 years ago