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Marysya12 [62]
3 years ago
13

Solve the following question algebraically. SOMEONE PLEASE HELP ME! ILL MARK THE BRAINLIEST ANSWER!!I’VE TRIED TO FIGURE IT OUT

ON MY OWN BUT IT WAS WRONG!!!!

Mathematics
1 answer:
Delicious77 [7]3 years ago
8 0

Solve for x. Isolate the x. Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS.

First, multiply 3 to both sides

6(3) = ((x + 2)/3)(3)

18 = x + 2

Finally, isolate the x. Subtract 2 from both sides

18 (-2) = x + 2 (-2)

x = 18 - 2

x = 16

16 is your answer for x

hope this helps

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Answer:

See Below

Step-by-step explanation

1) subtract 4 from both sides

You have the answer

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2 years ago
Solve by completing the square. x^2 + 4x – 1= 0
NNADVOKAT [17]

Answer:

x = -2 ± sqrt(5)

Step-by-step explanation:

x^2 + 4x – 1= 0

Add 1 to each side

x^2 + 4x – 1+1= 0+1

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Take the coefficient of x

4

Divide by 2

4/2 =2

Square it

2^2=4

Add it to each side

x^2 +4x+4=1+4

(x+2)^2 = 5

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Subtract 2 from each side

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8 0
3 years ago
Evaluate the expression 9-3÷1/3+1 <br>A. 9 <br>B. 3 <br>C. 1 <br>D. 19​
ikadub [295]
PEMDAS
P- none
E- none
M- none
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7 0
2 years ago
Let L be the line parametrized by x = 2 + 2t, y = 3t, z = −1 − t. (a) Find a linear equation for the plane that is perpendicular
Elden [556K]

Answer:

Step-by-step explanation:

Given that L is a line parametrized by

x = 2 + 2t, y = 3t, z = −1 − t

The plane perpendicular to the line will have normal as this line and hence direction ratios of normal would be coefficient of t in x,y,z

i.e. (2,3,-1)

So equation of the plane would be of the form

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Use the fact that the plane passes through (2,0,-1) and hence this point will satisfy this equation.

2(2)+3(0)-(-1) =K\\K =5

So equation is

2x+3y-z =5

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i.e. same point given.

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