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2 years ago

9

What is the slope of the equation y – 3 = –4(x – 5)?

1 answer:

Gemiola [76]2 years ago

8 0

Find the x and y intercepts or any other 2 points. (I advise you finding the y intercept though.)

then calculate for slope using:
y(2) - y(1)
________
x(2) - x(1)

The (2) and (1) are not exponents they are small numbers at the bottom of the numbers.

The slope is: 4

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2 years ago

Write the equation of the line parallel to 2x - 6y = 12 that passes through the point ( -1, 4).

vlada-n [284]

Answer:

y = 1/3x + 4 1/3

Step-by-step explanation:

2x - 6y = 12

First find the slope

Solve for y

Subtract 2x

2x - 6y -2x= 12-2x

-6y = -2x+12

Divide by -6

-6y/-6 =-2x/-6 +12/-6

y = 1/3 x -2

The slope is 1/3

Parallel lines have the same slope

Using the slope intercept form y = mx+b where m is the slope and b is the y intercept

y = 1/3x +b

And substituting in the point

4 = 1/3 (-1) +b

4 = -1/3 +b

Add 1/3 to each side

4 + 1/3 = b

y = 1/3x + 4 1/3

4 0

2 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In 2x - In(3x - 1) = 0

LenKa [72]

Answer:

$\frac{-1}{3x^2-x}$

Step-by-step explanation:

If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)= $\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}$

h(x)=In(3x - 1) then h'(x)= $\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}$

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3 years ago

A car traveling 118 miles uses 4 gallons of gas. At the rate, how many gallons will it use traveling 227 miles?

sashaice [31]

Answer:

b.7.69

Step-by-step explanation:

188/4=29.5

29.5 gal per mile

227/29.5=7.69

8 0

2 years ago

Y = 4x+1 y=x^2 + 2x - 2 A.6,2 B. 2,6 C. -6,2 D. -6,-2

natta225 [31]

Answer:

The solution is (3,13) and (-1,-3). So none of the mentioned options is correct.

Step-by-step explanation:

Given that

$y = 4x + 1 (i)$

$y = x^2 + 2x -2 (ii)$

Now, by susbstituting the value of 'y' from equation i to equation ii, we get

$4x+1=x^2 + 2x -2$

$0=x^2 + 2x -2-4x-1$

$0=x^2 + (2x -4x) +(-2-1)$

$0=x^2 + (-2x) +(-3)$

$0=x^2 -2x -3$

$x^2 -2x -3 = 0 (iii)$

Now by factorization, equation iii can be written as

$x^2 -3x +x -3 = 0$

$x(x -3) + 1(x -3) = 0$

$(x -3)(x +1) = 0$

x = 3 and x = -1

By putting the values of x in equation i, we get

y = 4(3) + 1

y = 12 +1

y = 13

and

y = 4(-1) + 1

y = -4 +1

y = -3

Therefore, the solution is (3,13) and (-1,-3). So none of the mentioned options is correct.

7 0

2 years ago