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lord [1]
3 years ago
11

The graph of a linear equation passes through the points (-4,1) and (4,6) HELP!

Mathematics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

a) (0, 3.5)

b) (12, 11)

Step-by-step explanation:

The points through which the graph of the linear equation passes are;

(-4, 1) and (4, 6)

The slope of the equation of a straight line passing through points (x₁, y₁), and (x₂, y₂), 'm', is given as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

∴ The slope of the equation, m = (6 - 1)/(4 - (-4)) = 5/8 = 0.625

The equation of th line in point and slope form is therefore;

y - 1 = 0.625·(x - (-4))

∴ y = 0.625·(x - (-4)) + 1 = 0.625·x + 3.5

y = 0.625·x + 3.5

Therefore;

a) When x = 0, y = 0.625 × 0 + 3.5 = 3.5

∴ The point (0, 3.5) is a solution

b) When x = 12, we have, y = 0.625 × 12 + 3.5 = 11

∴ The point (12, 11) is a solution

However;

c) When x = 8, we have, y = 0.625 × 8 + 3.5 = 8.5

∴ The point (8, 5) is not a solution

d) When x = -6, we have, y = 0.625 × (-6) + 3.5 = -0.25

∴ The point (-6, 0) is not a solution

Therefore;

The points which are solutions are;

(0, 3.5) and (12, 11)

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<h2><u>can use cross multiplication </u></h2>

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3 years ago
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Step-by-step explanation:

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2 years ago
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About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive
Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

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a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

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f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0
2 years ago
2+2 my question must be 20 characters long
geniusboy [140]
It is four. So two and two you add them together to be four. Good job. Your learning :)
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What operation should I use to solve -6x = -24. Addition, subtraction, multiplication,or division
olga_2 [115]
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