I need 9-12 please!!

Mathematics

1 answer:

11Alexandr11 [23.1K]2 years ago

8 0

9:
-q^2 - r^2 + 3s

-9^2 - -6^2 + 3(-20)

-9^2 = -9 * -9 = 81

81 - -6^2 + 3(-20)

-6^2 = -6 * -6 = 36

81 - 36 + 3(-20)

3(-20) = -60

81 - 36 + -60

81 + 36 = 117

117 + (-60)

117 - 60= 57

57

10:
3| x + y |^2 - (xy)^2

3| 3 + -5 |^2 - (3-5)^2

(3-5) = 2

3| 3 + -5 |^2 - 2^2

|3 + -5| = 8

3|8|^2 - 2^2

8 * 8 = 64

3(64) - 2^2

2^2 = 2*2 = 4

3(64) - 4

3*64 = 192

192 - 4 = 188

188

11:
2x^2 - 5xy - y^3

2(-3)^2 - 5(-3-2) - -2^3

(-3-2) = -5

2(-3)^2 - 5(-5) - -2^3

-3^2 = -3 * -3 = -9

2(-9) - 5(-5) - -2^3

2^3 = 2 * 2 * 2 = 8

2(-9) - 5(-5) - 8

18 - 5(-5) - 8

5(-5) = -25

18 - (-25) - 8

18 + 25 - 8

18 + 25 = 43

43 - 8 = 35

35

12: -a^2 + 7b^4 -2c^3

—4^2 + 7(-2)^4 - 2(-3)^3

4^2 + 7(-2)^4 - 2(-3)^3

4 * 4 = 16

16 + 7(-2)^4 - 2(-3)^3

-2 * -2 * -2 * -2 = 16

16 + 7(16) - 2(-3)^3

-3 * -3 * -3 = -27

16 + 7(16) - 2(-27)

7 * 16 = 112

16 + 112 - 2(-27)

2 * -27 = -54

16 + 112 - -54

16 + 112 + 54

112 + 16 = 128

128 + 54 = 182

182

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yanalaym [24]

<h3>Answer: 40</h3>

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Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

x/y = 2/1

1*x = y*2

x = 2y

Now use the segment addition postulate

JQ + QN = JN

x + y = 60

2y + y = 60

3y = 60

y = 60/3