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IrinaK [193]
3 years ago
12

I WILL MARK AS BRAINLIEST! Write an equation that represents each of the lines.

Mathematics
2 answers:
Ronch [10]3 years ago
5 0

Step-by-step explanation:

a. y = ¾x +6

b. y = 6/4 x -2

c. y = -2/4 x +3

d. y = -1/7 x -4

Karolina [17]3 years ago
3 0
Idc kid I’m sorry so sorry
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Simplify. √-64 A) -8 B) 8i C) -8i D) 32i
Vedmedyk [2.9K]

Answer:

C

Step-by-step explanation:

8i

√-64 = i√64 = 8i

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X^2 - 7x + 12
factors of 12      sum of the factors
1 * 12                13
2 * 6                    8
3 * 4                    7 
-1 * -12               -13
-2 * -6                  -8
-3 * -4                  -7
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4 years ago
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6 0
3 years ago
Determine the domain of the graph
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5 0
3 years ago
Find the coefficient of x of (1+2/x²) (2x-3/x)⁵
jenyasd209 [6]

First, we have

(1 + 2/x²) (2x - 3/x)⁵ = (2x - 3/x)⁵ + 2/x² (2x - 3/x)⁵

In the expansion of (2x - 3/x)⁵, there is no x² term. Each term takes the form

c (2x)^{5 - i} \left(-\dfrac3x\right)^i

where c is a binomial coefficient, and i is taken from the range {0, 1, 2, 3, 4, 5}. Looking at just the power of x in the product, we have

x^{5 - i} \left(\dfrac1x\right)^i = x^{5 - 2i}

and 5 - 2i = 1 only if i = 2. By the binomial theorem, this term is given by

\dbinom52 (2x)^{5-2} \left(-\dfrac3x\right)^2 = \dfrac{5!}{2!(5-2)!} \cdot 2^3 \cdot (-3)^2 x = 720x

In the other expansion, we have an additional factor of 1/x², so that any given them in the expansion contains a power of x of the form

\dfrac1{x^2} x^{5 - i} \left(\dfrac1x\right)^i = x^{3 - 2i}

and 3 - 2i = 1 only if x = 1. This term in the expansion is

\dfrac2{x^2} \dbinom51 (2x)^{5-1} \left(-\dfrac3x\right)^1 = \dfrac{5!}{1!(5-1)!} \cdot 2^5 \cdot (-3)^1 x = -480x

Then the coefficient of the x term in the whole expansion is 720 - 480 = 240.

8 0
3 years ago
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