1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
xxTIMURxx [149]
3 years ago
5

Write an equation of a line perpendicular to y=3x+5 that contains the points (-3,-4)

Mathematics
1 answer:
Vlad [161]3 years ago
5 0

The equation of line is:

y = - \frac{1}{3}x-5

Further explanation:

Given equation of line is:

y = 3x+5

The standard form of point-slope form is:

y = mx+b

Comparing the given equation with the standard form is:

m = 3

We know that product of slopes of two perpendicular lines is -1

Let m2 be the slope of line perpendicular to y = 3x+5

Then

3 * m_2 = -1\\m_2 = -\frac{1}{3}

Putting the value of slope in standard form

y = -\frac{1}{3}x + b

To find the value of b, putting (-3,-4) in equation

-4 = -\frac{1}{3}(-3) + b\\-4 = 1 + b\\b = -4-1\\b = -5

Putting the values of slope and b in equation

y = - \frac{1}{3}x-5

Keywords: Slope, Point-intercept form

Learn more about perpendicular lines at:

  • brainly.com/question/1332667
  • brainly.com/question/13096001

#LearnwithBrainly

You might be interested in
How many liters each of a 35% acid solution and a 80% acid solution must be used to produce 60 liters of a 65% acid solution? (R
serg [7]

Solving a system of equations we will see that we need to use <u>40 liters of the 80% acid solution</u>, and the other <u>20 liters are of the 35% acid solution</u>.

<h3>How many liters of each solution do we need to use?</h3>

First, we need to define the variables:

  • x = liters of the 35% acid used.
  • y = liters of the 80% acid used.

We know that we want to produce 60 liters of 65% acid, then we have the system of equations:

x + y = 60

x*0.35 + y*0.80 = 60*0.65

(in the second equation we wrote the percentages in decimal form).

To solve this we need to isolate one of the variables in one equation and then replace it in other one, isolating x we get:

x = 60 - y

Replacing that in the other equation:

(60 - y)*0.35 + y*0.80 = 60*0.65

y*(0.80 - 0.35) = 60*(0.65 - 0.35)

y*0.45 = 60*0.30

y = 60*0.30/0.45 = 40

So we need to use <u>40 liters of the 80% acid solution</u>, and the other <u>20 liters are of the 35% acid solution</u>.

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

7 0
2 years ago
How do u put 72 7/10 into decimal?
NNADVOKAT [17]

Answer:

72.7

Step-by-step explanation:

First multiply 72 by 10

because 72 is 72 10s, making it a whole number.

then add 720 (72 times 10) to 7.

then all of that over 10.

then just divide.

727/10

72.7

this is what you do for all mixed numbers like that one.

7 0
2 years ago
Read 2 more answers
In the triangle below, what is the tangent of 45°?
Lana71 [14]

Answer:

C

Step-by-step explanation:

tan45° = \frac{opposite}{adjacent} = \frac{1}{1} = 1

6 0
3 years ago
Read 2 more answers
What is 150% of 200. Show work. ASAP
zalisa [80]
150 % of 200

150 / 100 =  1.5

1.5 * 200 = 300

hope this helps!.
8 0
3 years ago
Read 2 more answers
Solve 48-x=10x+12-4x
MissTica
The answer is 5 and 1/7
6 0
3 years ago
Read 2 more answers
Other questions:
  • 23/20 or 110% which is greater?
    12·1 answer
  • Please help with synthetic division:
    13·1 answer
  • Factorise 32 x cube minus 8 y square​
    11·1 answer
  • How do i simplify this
    14·1 answer
  • Solve for y, then graph equation <br><br> Y+2x=3
    13·1 answer
  • What is the answer to c-(-12)=16
    5·1 answer
  • Please help, due tomorrow
    13·2 answers
  • several friends each had 2/5 of a bag of peanuts left over from game. they realized that they could have bought 2 fewer bags. Ho
    15·1 answer
  • Answer quick Question 2 of 5
    7·2 answers
  • PLEASE HELP! ( no random links ) correct answers please. Find median
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!