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2 years ago

5

write a equation for the vertical and horizonal lines passing through the points (4,1) I know x=4 and y=1 but what eles do I do?

1 answer:

Harlamova29_29 [7]2 years ago

6 0

Well you solve the equation then find out what you dealing with

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Solution set is(answer)................

kirill [66]

<h3>The solution is x = 27</h3>

To undo a log, you raise both sides as exponents. This will cancel out the log on the left side. On the right side you'll have 6^2 turn into 36.

So you should end up with x+9 = 36 after you perform this step. Afterwards, subtract 9 from both sides to end up with x = 27.

4 0

2 years ago

What is the answer to #14?

scZoUnD [109]

It think it’s 6x...!

7 0

3 years ago

Determine the equation of the line that has an x-intercept of 6 and passes through the point (6,4)

butalik [34]

Putting a = 6 and b= 3/2 in eqn. (1) the required equation is:-

x /6 + 2.y/3 = 1. or, x + 4y = 6. Answer.

5 0

2 years ago

Read 2 more answers

0.35, 0.68, 0.20, 0.31<br><br> Order the numbers greatest to least.

Artyom0805 [142]

Answer:0.68,0.35,0.31,0.20

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16

Masja [62]

Answer: $\frac{x-4}{2x-2}$

Step-by-step explanation:

$\frac{2x^2+5x+3}{x^2-3x-4}$ divided by $\frac{4x^2+2x-6}{x^2-8x+16}$ is the same thing as multiplying $\frac{2x^2+5x+3}{x^2-3x-4}$ by $\frac{x^2-8x+16}{4x^2+2x-6}$ .

The equation we get is:

$\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}$

With this equation, we can factor each equation:

$\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}$

We can cancel out like terms since they would be dividing each other:

$\frac{x-4}{2x-2}$

8 0

2 years ago