Find the equation of the line that is parallel to the given line and passes through (9,1) y=2/3x -2

1 answer:

8 0

When the lines are parallel, the slopes of these lines are equal.
For a new line,
y = (2/3)x + b.

To find b, we need substitute x and y with values of the point (9,1).
x=9
y=1

1 = (2/3)*9 + b
1 = 6 + b
b=-5

The equation of the line is y = (2/3)x - 5.

You might be interested in

What's is the answer to 12% of 91

MakcuM [25]

First, turn 12 percent into decimal which is .12
Of = multiple
So now you times .12 times 91 which is 10.92
Answer 10.92
Hope this help

8 0

2 years ago

Which of the following is a pair of vertical angles? <br><br> EXAMPLE: 1 and 4 , 1 and 5

tresset_1 [31]

5 and 7
1 and 3
4 and 2
8 and 65

6 0

2 years ago

Read 2 more answers

(1,2) and (7,9) I need to find the slope of that

ehidna [41]

Answer:

7/6 is your slope,,,,,

8 0

3 years ago

Read 2 more answers

What is the solution of the inequality below? <br> y-1≥0

irinina [24]

Answer:

y≥1

Step-by-step explanation:

hello :

y-1≥0

y-1+1≥0+1

y≥1

3 0

2 years ago

Make 'h' as the subject. Remember to rationalize the denominator while solving.

Sliva [168]

Hello there BeGuiLE !

To solve your question, we must first bring 'h' to one side of the equation. Let's bring all the terms with the variable 'h' to the left side of the equation. So,
$\large{\sf\sqrt{ 3 } h = 1.6+h}$
→ $\large{\sf\sqrt{3}h-h=1.6 }$

Now, let's combine the 2 terms containing the variable 'h'.
$\large{\sf\sqrt{3}h-h=1.6 }$
→ $\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
$\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$
→ $\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
$\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }$
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }$
→ $\boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}$

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
$\large{\sf\:h=0.8(\sqrt{3}+1)}$
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ $\large{\sf\:h=0.8(1.73+1)}$
→ $\large{\sf\:h=0.8(2.73)}$
→ $\boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}$

I hope this will help you.