Find an equation of the line parallel to y= 1/3x +1 and that passes through the point (3,5)

1 answer:

6 0

So in order to find the line you’ll have to submit the point (3,5) and the slope into the point slope formula : y-y1= m(x-x1)

Y-5= 1/3(x-3)

Solve(turn to y intercept form):

Y-5= 1/3x -1

You might be interested in

How would I do this? Solve for q. –10 + 10q = 9q

DerKrebs [107]

Answer:

q=10

Step-by-step explanation:

-10+10q=9q

-10=-q

q=10

4 0

3 years ago

Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of z0 = 5i.

Agata [3.3K]

3 because the 2z minus the 3 gives you 1 rounded

4 0

3 years ago

Read 2 more answers

Mr. Seidel filled up his 14 gallon gas tank for 29.26. What was the price per gallon?

mr_godi [17]

It would be $2.09 if you are dividing

6 0

3 years ago

Read 2 more answers

(20 points)<br> Question from Similarity

Mariana [72]

Answer:

45 is the anwser

Step-by-step explanation:

thank you

7 0

3 years ago

What is the area of a trapezoid ABCD with bases AB and CD ,

Mrrafil [7]

check the picture below on the top side.

we know that x = 4 = b, therefore, using the 30-60-90 rule, h = 4√3, and DC = 4+8+4 = 16.

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\[-0.5em]
\hrulefill\\
a=8\\
b=\stackrel{DC}{16}\\
h=4\sqrt{3}
\end{cases}\implies A=\cfrac{4\sqrt{3}(8+16)}{2}
\\\\\\
A=2\sqrt{3}(24)\implies \boxed{A=48\sqrt{3}}$

now, check the picture below on the bottom side.

since we know x = 9, then b = 9, therefore DC = 9+6+9 = 24, and h = b = 9.

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\[-0.5em]
\hrulefill\\
a=6\\
b=\stackrel{DC}{24}\\
h=9
\end{cases}\implies A=\cfrac{9(6+24)}{2}
\\\\\\
A=\cfrac{9(30)}{2}\implies \boxed{A=135}$

7 0

4 years ago