JK has endpoints at J(9, 9) and K(1, 3). Find the midpoint M of JK.

1 answer:

3 0

The midpoint of JK will be the points in between the x values and the y values.

There is an easy midpoint formula that goes as follows:

M= ((x1+x2)/2, (y1+y2)/2)

Now, we go 9+1 which is 10 and divide by 2 getting 5.

For the y values we do the same- 9+3 is 12 divided by 2 is 6.

The midpoint of JK is (5,6).

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Write the recursive formula for 5, 15, 30, 50, . . . . . .

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A recursive sequence is a sequence in which terms are defined using one or more previous terms that are given.
If you know the term n of an arithmetic sequence and you know the common difference, d, you can find the term (n + 1) by tossing the recursive formula to n + 1 = a n + d

We have then:
5, 15, 30, 50
an = (5/2) * (n) * (n + 1)
Answer:
the recursive formula for 5, 15, 30, 50, is:
an = (5/2) * (n) * (n + 1)

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3 years ago

Complete the steps to factor the polynomial. One root of f(x) = x³ + x² - 22x - 40 is 5. If 5 is a root of the function, then is

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If one root of a cubic is known (and verified), then we can do a synthetic division and factor the remaining quadratic expression.

f(x)=x^3+x^2-22x-40
f(5)=125+25-22(5)-40=0 so x-5 is a factor

Now do the synthetic division

5 | +1 +1 -22 -40
--------------------
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The resulting quadratic is x^2+6x+8 which can be factored into
(x+2)(x+4)

The complete factorization is therefore
f(x)=<span>x³ + x² - 22x - 40 = (x-5)(x+2)(x+4)</span>

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3 years ago

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If f(x) = 2x ^ 2 - 5 and g(x) = x ^ 2 - 4x - 8 find (f - g)(x)

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Answer: (f-g)(x) is equivalent to f(x) - g(x)

So that would be (2x²-5) - (x²-4x-8) = 2x² - x² + 4x + 8 - 5 = x² + 4x + 3

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