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2 years ago

Find the coordinate of the midpoint M given the two endpoints

Mathematics

1 answer:

Rasek [7]2 years ago

8 0

The midpoint would be (7, 5)

The midpoint of any two points is simply the average of the x points and the average of the y points. So we'll start by averaging the 2 x points together. To take an average, we add the two numbers together and divide by 2.

X VALUE

4 + 10 = 14

14/2 = 7

Now do the same for y

Y VALUE

22 + -12 = 10

10/2 = 5

Using those answers we can create the ordered pair (7, 5)

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The polynomial of degree 3, P(x), has a root of multiplicity 2 at x = 3 and a root of multiplicity 1 at x = -2. The y-intercept

Maru [420]

Answer:

$p(x) = - 0.7(x - 3)^{2} (x + 2)$

Step-by-step explanation:

We have that the polynomial, has a root of multiplicity 2 at x = 3 and a root of multiplicity 1 at x = -2.

This means the factored form of the polynomial will be.

$p(x) = a(x - 3)^{2} (x + 2)$

Also, it was given that, the y-intercept is y=-12.6.

This implies that:

$- 12.6= a(0 - 3)^{2} (0 + 2)$

$- 12.6= a(- 3)^{2} (2)$

$- 12.6= 18a$

Divide both sides by 18;

$a = \frac{ - 12.6}{18}$

$a= - 0.7$

Therefore the polynomial is

$p(x) = - 0.7(x - 3)^{2} (x + 2)$

7 0

3 years ago

Please help i wanna move to the next grade,10 points

AfilCa [17]

Answer:

C -1

Step-by-step explanation:

$- 3x + 2 < 5$

$5 - 2 = 3$

$- 3x \div - 3$

$3 \div - 3 = - 1$

$x = - 1$

4 0

2 years ago

Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C

netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> $|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}$ </span></span>

Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.

$|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5$

$|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6$

$|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4$

$|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \ \sqrt{25}=5$

$|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4$

Answer: B and D

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Musya8 [376]

A. 97,280 feet

B. 18.42mph

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