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1 year ago

What is the equation of the line that passes through the point (8,-4) and has a slope of 3?

2 answers:

8 0

In order to solve it you insert the point in slope from

the equation is
Y=Mx+b

Plug it in as

-4=3(8)+b

-4=24+b

Then subtract 24

B=-28

So

The answer is

Y=3x-28

sineoko [7]1 year ago

8 0

Y-y1=m(x-x1)

(8, -4) 8 is x1 and -4 is y1
so y- (-4) = 3 (x -8)
make 4 a positive then y + 4 = 3x - 24
subtract 4 from 24 then y=3x-20

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Find the slope of the line between the points (1, 4) and (-1,6)

Oduvanchick [21]

Answer:

slope = - 1

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 4) and (x₂, y₂ ) = (- 1, 6)

m = $\frac{6-4}{-1-1}$ = $\frac{2}{-2}$ = - 1

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2 years ago

The area of a particular rectangle is 72. If the length of the rectangle is twice

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

Suppose the width is "a". Then the length is 2a.

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2 years ago

1/x + 1/3x =6 how do you solve this equation for x

Elodia [21]

Answer:

x = 2/9 (or = 0.222)

Step-by-step explanation:

1/x + 1/(3x) =6 (multiply both sides by 3x)

(1/x) (3x) + [1/(3x)](3x) =6 (3x)

3 + 1 = 18x

4 = 18x

x = 4/18

x = 2/9 (or = 0.222)

3 0

2 years ago

Given: 2x^2+3x-5, g(x)=x+9. Find: (fg)(3)

Svetradugi [14.3K]

Answer:

264

Step-by-step explanation:

$f(x)=2x^2+3x-5$

$g(x)=x+9$

$(fg)(x)=f(x) \times g(x)$

$\implies (fg)(x)=(2x^2+3x-5)(x+9)$

$(fg)(3)=(2(3)^2+3(3)-5)(3+9)$

$\implies (fg)(3)=22 \times 12$

$\implies (fg)(3)=264$

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1 year ago

What is the solution to the compound inequality? –6 ≤ 2x + 6 ≤ 6

maw [93]

∈ ≤ ≥

-6 ≤ 2x + 6 ≤ 6 ( seperate the two inequalities)

2x + 6 ≥ - 6

2x + 6 ≤ 6 (solve the inequalities)

x ≥ -6

x ≤ 0 (find the intersection)

solution is : x ∈ { -6, 0 }

7 0

3 years ago