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Romashka [77]
3 years ago
15

What is the equation of this line Y=1/4x Y=4x Y=-1/4x Y= -4x

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0
I know this is really late but the answer is y=-1/4x
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What is 76,808 rounded to the nearest 100?
Neporo4naja [7]

Answer:

76,800

Step-by-step explanation:

This is because 808 is closer to 800

5 0
3 years ago
Assume m<2 = 35°. Find the measure of all of the other
soldi70 [24.7K]

  1. m<1 = 145 | Supplementary
  2. m<3 = 35 | Vertical
  3. m<4 = 145 | Supplementary
  4. m<5 = 145 | (With Angle 4) If parallel then alternate interior angles congruent
  5. m<6 = 35 | (With Angle 5) Supplementary
  6. m<7 = 35 | (With Angle 6) Vertical
  7. m<8 = 145 | (With Angle 7) Supplementary

Hope it helps <3

(If it does, maybe brainliest :) Need one more for rank up)

6 0
3 years ago
Read 2 more answers
Use the figure to find the measure of angle 2.
ale4655 [162]

Answer:

∠ 2 = 70°

Step-by-step explanation:

110° and ∠ 1 are corresponding angles and congruent, thus

∠ 1 = 110°

∠ 1 and ∠ 2 are adjacent angles and are supplementary, thus

∠ 2 = 180° - ∠ 1 = 180° - 110° = 70°

7 0
3 years ago
The coordinates R(1, -3), S(3, -1) T(5, -7) form what type of polygon?
Kobotan [32]
<span>The coordinates R(1, -3), S(3, -1) T(5, -7) form a triangle. A triangle is a polygon that has 3 sides and all the angles add up to 180</span>°. 
8 0
2 years ago
Determine the temperature of 2.6 moles of gas contained in a 5.00-L vessel at a pressure of 1.2atm.
strojnjashka [21]

Answer:

28.108 K.

Step-by-step explanation:

Given: Pressure (P)= 1.2atm

           Number of moles (n)=  2.6 moles

           Volume (V)= 5.00-L

Now finding the temperature (T).

Formula; T= \frac{P\times V}{n\times R}

R is a constant factor which makes other factors work together.

There is a numerical value for R which we use is 0.0821 \times \frac{L.atm}{mole.K}

∴ Temperature (T)= \frac{1.2\times 5}{2.6\times 0.0821 \frac{L.atm}{mol.K} }

⇒ Temperature (T)= \frac{6}{0.21346} = 28.1083\ K

∴ Temperature is 28.108 K

8 0
2 years ago
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