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taurus [48]
4 years ago
14

Simplify 2/3[12v-15]

Mathematics
1 answer:
pychu [463]4 years ago
5 0
<span>I believe that  2*(4v-5) is the right answer</span>
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Isaac wants the equation below to have no solution when the missing number is placed in the box.
MatroZZZ [7]

Answer: 4

Step-by-step explanation: To have an equation with no solution you have to get something like 0=1 meaning you have to remove all the variables. In this equation you would distribute to get xy+2y+2x=2x+12+4x where y is the missing number. then combining like terms you get xy+2x+2y=6x+12. Subtracting 2x from both sides you get xy+2y=4x+12. Now to get the variables gone y has to be 4 so that the variables to cancel out. By plugging in 4 you can see this works; 4x+8=4x+12 and by subtracting 4x from both sides you get 8=12 which is not true meaning there is no solution.

8 0
3 years ago
Read 2 more answers
Solve for substitution 8x+2y=13 and 4x+y=11
damaskus [11]
Solving for X:
1) y=-8x+6.5   y=-4x+11
2) -8x+6.5=-4x+11
3) -8x=-4x+11-6.5        (Subtract 6.5 from both sides)
4) -8x+4x=11-6.5         (Add 4x to both sides)
5)-4x=4.5
5)x=-1.125

Solving for Y:
y=-8(-1.125)+6.5
y=9+6.5
y=15.5

y=-4(-1.125)+11
y=4.5+11
y=15.5

x=-1.125
y=15.5
or
(-1.125,15.5)
7 0
4 years ago
What is the answer for N-6/3=4
konstantin123 [22]
N=6 Is the answer to your question
4 0
4 years ago
Jessie wants to find out what type of music her fellow students enjoy the most. She asks her classmates: "What is your favorite
iogann1982 [59]
Country cause it's good
5 0
3 years ago
PROVE THAT:<br><br>cos 20° - sin 20° = ​​ \sqrt{2}sin25°<br><br>​
Lemur [1.5K]

Answer:

See below.

Step-by-step explanation:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

First, use the co-function identity:

\sin(90-x)=\cos(x)

We can turn the second term into cosine:

\sin(20)=\sin(90-70)=\cos(70)

Substitute:

\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Now, use the sum to product formulas. We will use the following:

\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})

Substitute:

\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70)  =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)

Use the even-odd identity:

\sin(-x)=-\sin(x)

Therefore:

\cos(20)-\cos(70)=-\sqrt{2}\sin(-25)\\\cos(20)-\cos(70)=-\sqrt{2}\cdot-\sin(25)\\\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Replace the second term with the original term:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

Proof complete.

4 0
4 years ago
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