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2 years ago

What is the equation of the line that passes through the point (-7, 2) and has a slope of -1?

Mathematics

1 answer:

lorasvet [3.4K]2 years ago

3 0

Answer: y= -1x - 5

Step-by-step explanation:

The equation has to be in slope-intercept form which uses the formula

y=mx+b where m is the slope and b is the y-intercept.

So we given the slope and an x and y coordinate so we will plot those numbers into the formula and solve for the y intercept to write the equation.

the x coordinate is -7 and the y coordinate is 2

2 = -1(-7) + b

2 = 7 + b

-7 -7

b = -5 the y intercept is -5

Now the equation will be y= -1x - 5

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You get in an elevator and go down 8 floors. You exit on the 16th floor. Write and solve an equation to find what floor you got

disa [49]

So floor that you started is x
go down 8 means -8
end on 16

so
x-8=16
add 8
x=24

started on 24

3 0

3 years ago

Find the slope of a line parallel to the line that crosses ( 3 , − 2 ) and the y -intercept is ( 0 , − 3 ) . I finally figured o

mamaluj [8]

Answer:

y = 1/3x - 3

Step-by-step explanation:

We can find the equation of the line, by finding the slope and combine with our y-intercept (-3).

We need to use the slope formula to find the slope.

Thus, we have (-3 - (-2)) / 0 - 3 = -1/-3 = 1/3

So our equation is y = -1/3x - 3

8 0

1 year ago

what digit is in the ten thousands place 1.03098 which is the correct answer I know it's one of the zeros I believe which one is

elena-14-01-66 [18.8K]

Hello there

Actually, the correct answer is the 9

the first 0 is in the tenths spot, the 3 is in the hundredths, the second zero is in the thousandths, and the 9 is in the ten thousandths, while the 8 is in the hundred thousandths.

Therefore, the 9 is in the ten thousandths place

I hope this helped ^^

4 0

2 years ago

What is .08333333333 as a fraction ?

madam [21]

<span>.0833333333 with repeating 3 is (2/24) or (1/12) <- There's you're answer.

<em>(Just put this up here since I answered it in a comment)</em></span>

5 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

2 years ago