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GarryVolchara [31]
3 years ago
12

How do I solve the equation 9+t/12=-3?

Mathematics
2 answers:
Alex17521 [72]3 years ago
8 0

Answer:

\Huge\boxed{\mathsf{\Rightarrow T=-144}}

Step-by-step explanation:

\Large\boxed{\textnormal{SUBJECT: MATH}}}

<h3><u /></h3><h3><u>TO SOLVE:</u></h3>

Isolate by the t on one side of the equation.

<h3><u>SOLUTIONS:</u></h3>

First, subtract 9 from both sides.

\displaystyle \mathsf{9+\frac{t}{12}-9=-3-9 }

Solve.

\displaystyle \mathsf{\Rightarrow \frac{t}{12}=-12 }

Then, multiply by 12 from both sides.

\displaystyle \mathsf{\frac{12t}{12}=12(-12) }}

Solve.

Multiply.

\displaystyle \mathsf{12*(-12)=\boxed{\mathsf{\rightarrow -144}}}

\Rightarrow \Large\boxed{\mathsf{T=-144}}

The correct answer is t=-144.

Bad White [126]3 years ago
7 0

Answer:

t=-144

Step-by-step explanation:

t in (-oo:+oo)

t/12+9 = -3 // + 3

t/12+3+9 = 0

1/12*t+12 = 0 // - 12

1/12*t = -12 // : 1/12

t = -12/1/12

t = -144

t = -144

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Answer:

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Step-by-step explanation:

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4 0
4 years ago
Help plsss!!!! Thnk you!!!!
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Step-by-step explanation:

General quadratic equation: y = ax² + bx + c.

When x = 0, y = -3.

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We now have y = ax² + bx - 3.

When x = 2, y = -4.

=> (-4) = a(2)² + b(2) - 3, 4a + 2b - 3 = -4,

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Our 1st system of equations is 4a + 2b = -1.

When x = 4, y = -3.

=> (-3) = a(4)² + b(4) - 3, 16a + 4b - 3 = -3,

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Our 2nd system of equations is 4a + b = 0.

Now we solve the system of equations:

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=> b = -1

Hence 4a + (-1) = 0, 4a = 1, a = 1/4. (B)

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3 years ago
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