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12345 [234]
3 years ago
8

Solve each equation for m. Then find the value of m for each value of n.

Mathematics
1 answer:
Misha Larkins [42]3 years ago
5 0
If n = –2, then m is equal to 13;

If n = 0, then m is equal to 7;

If n = 1, then m is equal to 4
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Find the distance between the two points. (-1, 6), (2, 8) Round to the nearest hundredth
MakcuM [25]

Answer:

d = \sqrt{13} or d = 3.61

Step-by-step explanation:

d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d= \sqrt{(2-(-1))^2+(8-6)^2}\\d= \sqrt{(3)^2+(2)^2}\\d= \sqrt{9+4}\\d= \sqrt{13}

5 0
3 years ago
Given the following inverse variation find the missing value: (5,-8) (20, y) find the missing value
forsale [732]
This answer is pretty simple. you see (5,-8) and (20,y). well the 20 is 4 times the 5 on the x value so multiply the y value by 4 to get your answer which Y=-32 in the 2nd corrdenants.
6 0
3 years ago
Rewrite the polynomial in the form ax^2+bx+c and then identify the values of a, b, and c. 6+9x^2
Ulleksa [173]

Answer:

see below

Step-by-step explanation:

6+9x^2

Rewriting in decreasing powers of x

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ax^2+bx+c

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c = 6

5 0
3 years ago
X=a<br> There were 120 problems.'<br> You got two-fifths wrong.<br> How many did you get right?
swat32

120 x 2 = 240

240/5 = 48

120 - 48 = 72

Answer is 72.

You got 72 correct.

7 0
3 years ago
What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 −
Anna007 [38]

Answer:

b.

Step-by-step explanation:

We have to look at sign changes in f(x) to determine the possible positive real roots.

f(x)=-3x^4-5x^3-x^2-8x+4

There is only one sign change here, between the -8x and the +4.  So that means there is only 1 possible real positive root.

Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

f(-x)=-3x^4+5x^3-x^2+8x+4

There are 3 sign changes here.  That means there are either 3 negative roots or 3-2 = 1 negative root.  So we have:

1 positive

3 or 1 negative

We need to pair them up now with all the possible combinations.

If we have 1 positive and 1 negative, we have to have 2 imaginary

If we have 1 positive and 3 negative, we have to have 0 imaginary

Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial.  This is a 4th degree polynomial, so we will have 4 roots.

7 0
3 years ago
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