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bezimeni [28]
2 years ago
8

Which linear inequality could represent the given table of values? y < –2x + 3 y ≤ –2x + 3 y > –One-halfx – 3 y ≤ –One-hal

fx – 3
Mathematics
1 answer:
Margarita [4]2 years ago
8 1

Answer:

< –2x + 3 y ≤ –2x + 3 y > hope this helps lol...,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

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Determine whether the vectors u and v are parallel, orthogonal, or neither.
kicyunya [14]

Answer:

B.) Orthogonal

Step-by-step explanation:

Two vectors u and v whose dot product is u·v=0 are said to be orthogonal

u = <6, -2>, v = <2, 6>

u·v = u1*v1 + u2*v2

          6*2 + -2 * 6

        =12 -12

         =0

4 0
3 years ago
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Okay here are more math problems.
Vinil7 [7]

Answer:

the inequality would be, b/3<=16

Step-by-step explanation:

6 0
2 years ago
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Leo the Rabbit
Ghella [55]
23 I think idk I try
3 0
2 years ago
Can someone help me with this? its solving systems by substitution
yKpoI14uk [10]
Answer: x = 1 and y = 2

Given:
y = -3x + 5
5x - 4y = -3

Since we are given y, let’s sub it in the second equation.

5x - 4(-3x + 5) = -3
5x + 12x - 20 = -3
17x = 17
x = 1

After finding x, we can now find y.

5(1) - 4y = -3
-4y = -8
y = 2

Checking:

y = -3x + 5
2 = -3(1) + 5
2 = 2


5x - 4y = -3
5(1) - 4(2) = -3
-3 = -3
8 0
2 years ago
In january of one winter, 3/10 of a wood pile was used in a wood stove. In February, 2/5 of the wood pile was used. What part of
arsen [322]
2/5 is equivalent to 4/10 so at the end of February 4/10 was used up
And at the end of January, 3/10 was used
so 4/10 + 3/10 = 7/10

At the end of February, 7/10 of the wood was used
6 0
3 years ago
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