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Murrr4er [49]
4 years ago
11

Which variable is most efficient to isolate?

Mathematics
2 answers:
mezya [45]4 years ago
7 0

y in the first equation because y is already by itself. All you have to do is move the -3x to the other side.

Tems11 [23]4 years ago
4 0

The <u>y in the first equation</u> because all you have to do is add 3x on both sides, and then y is by itself.

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Find the gradient of the line passing through the points (– 2,– 4) and (3,5).
Nataliya [291]
Answer is 9/5
You can count the rise/run on the graph or use slope formula (y2-y1) over (x2 -x1)

5-(-4) over 3-(-2) = 9/5
6 0
2 years ago
Read 2 more answers
SIMPLIFY THE FOLLOWING EXPRESSION: given that s ≠ 0, and write your answer using only positive exponents. Thanks in advance!!!
Umnica [9.8K]

Answer:

4s¹²/(9t⁴)

Step-by-step explanation:

<u>Solving in steps:</u>

  • (3s⁻⁴t⁷s⁰)⁻²(-2s²t⁵)² =
  • 3⁻²s⁻⁴ˣ⁻²t⁷ˣ⁻²(-2)²s²ˣ²t⁵ˣ² =
  • 1/9×s⁸t⁻¹⁴(4)s⁴t¹⁰ =
  • 4/9 ×s⁸⁺⁴t⁻¹⁴⁺¹⁰ =
  • 4/9 ×s¹²t⁻⁴=
  • 4s¹²/(9t⁴)
3 0
3 years ago
4(3x+5)= -4
NeX [460]

Answer/explanation

4(3x+5)=-4

You will distribute the four to every term in the parenthesis,

12x+20=-4

Then you will combine like terms

12x+20=-4

     -20  -20

12x=-24

Then you will divide 12 to isolate x

\frac{12x}{-24} =\frac{-24}{-24}

Obviously, -24 cancels out and you would get x=-2

4 0
4 years ago
In some gymnastics meets, the score given to a gymnast is the mean of the judges’ scores after the highest and lowest scores hav
marishachu [46]

Answer:

her score is lower after removing the highest and lowest value

just took the test

6 0
3 years ago
A boat travel 312 miles each way downstream and back. The trip downstream took 13 hours. the trip back took 26 hours. What is th
zaharov [31]
Remember that Speed= \frac{distance}{time}. Since the distance of the complete trip is 312 miles, the distance of each way is \frac{312}{2} =156 miles. 
Let  S _{b} be the speed of the boat in still water and S_{c} the speed of the current.

Now, for the downstream trip the boat is traveling with the current, so:
S_{b} +S_{c} = \frac{156}{13}
S_{b} +S_{c} =12 this will be our equation (1)

For the trip back the boat is traveling against the current, so:
S _{b} -S_{c} = \frac{156}{26}
S_{b} -S_{c} =6 this will be our equation (2)
 
Next, lets add equation (1) and equation (2) to get rid of S _{c}:
\left \{ {{S_{b}+S_{c}  =12} \atop {+(S_{b}-S_{c}  =6})} \right.
2S_{b} =18
S _{b} = \frac{18}{2}
S _{b} =9

Finally, now that we know the speed of the boat in still water, lets replace that value in our equation (1) to find the speed of the current:
9+S_{c} =12
S_{c} =12-9
S_{c} =3

We can conclude that the speed of the boath in still water is 9\frac{mi}{h}, and the current of the stream is 3\frac{mi}{h}.


7 0
4 years ago
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