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Tomtit [17]
2 years ago
9

Find the slope of a parallel line to each given line y=-1/3x-4

Mathematics
2 answers:
Mumz [18]2 years ago
4 0
Slope of parallel line to this would be -1/3x
Eduardwww [97]2 years ago
4 0

Answer:

3

Step-by-step explanation:

3x-y =4

-y= 3x - subtract 3x from both sides

y= 3x - 4 multiply both sides by -1

the slope is 3

You might be interested in
3q+4+9=-14 <br> what is Q
BlackZzzverrR [31]

Hi! Your answer is q = -9

Please see an explanation for a better and clear understanding to your problem.

Any questions about my answer and explanation can be asked through comments! :)

Step-by-step explanation:

Since we want to solve for q-term. That means we are going to isolate q-term.

\huge{3q+4+9=-14}

We can add 4 and 9 together.

\huge{3q+13=-14}

Because we want to know the value of q. That means we have to isolate q-term by subtracting both sides by 13.

\huge{3q+13-13=-14-13}\\\huge{3q=-27}

We are reaching to the final step where we divide the whole equation by 3.

\huge{\frac{3q}{3}=-\frac{27}{3}}\\\huge{q=-9}

Finally, the solution for this equation is q = -9. But what if you are not certain or sure about the answer? Let's check it out!

To check the answer, simply substitute q = -9 in the equation.

\huge{3q+4+9=-14}\\\huge{3(-9)+13=-14}\\\huge{-27+13=-14}\\\huge{-14=-14}

Notice that the equation is true for q = -9. Hence, we can conclude that the solution for this equation is q = -9.

Hope this helps!

5 0
3 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
2 years ago
What is the standard form of (2+3i)(4-7i)/(1-I)
AlladinOne [14]
I believe in standard form this would be 31/2 + 27i/2
8 0
2 years ago
Read 2 more answers
Pls help me i need help
malfutka [58]

Answer:

4p^2-3

Step-by-step explanation:

8 0
2 years ago
36 is what percent of 200
Andrews [41]

Answer:

18

Step-by-step explanation:

I used a percentage calculator to get you the best answer possible. :) Have a nice day!

3 0
3 years ago
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