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Daniel [21]
2 years ago
14

Solve the equation. 2 + |t+ 6| = 12

Mathematics
2 answers:
Contact [7]2 years ago
8 0

Answer:

t = 4 or t = -16

Step-by-step explanation:

Let's solve your equation step-by-step.

2+|t+6|=12

|t+6|+2=12

Step 1: Add -2 to both sides.

|t+6|+2+−2=12+−2

|t+6|=10

Step 2: Solve Absolute Value.

|t+6|=10

We know eithert+6=10 or t+6=−10

t+6=10(Possibility 1)

t+6−6=10−6(Subtract 6 from both sides)

t=4

t+6=−10(Possibility 2)

t+6−6=−10−6(Subtract 6 from both sides)

t=−16

Yuri [45]2 years ago
5 0
T = 4 or -16 hope this helps
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3 years ago
Suzie made a mistake in the following problem. The mistake was made in Line ___. Only input the number of the first incorrect li
zalisa [80]

Answer:

Mistake in Line 2

Step-by-step explanation:

Line 1  

7(6) ÷ 5 + 42  

Line 2  

7(6) ÷ 5 + 16  

Line 3  

42 ÷ 5 + 16  

Line 4  

42 ÷ 21  

Line 5  

2

To simplify any expression we use order of operation

Line 1 is 7(6) ÷ 5 + 42  

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First we start with parenthesis

so we multiply 7(6) in line 2

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3 0
3 years ago
100 points gor correct answer Identify the base of a triangle in which h=(3x+15) ft and A=(15x+75) ft^2.
ivolga24 [154]

Answer:

b = 10

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the height )

Here h = 3x + 15 and A = 15x + 75, thus

\frac{1}{2} × b × (3x + 15) = 15x + 75

Multiply both sides by 2 to clear the fraction

b(3x + 15) = 30x + 150

Divide both sides by (3x + 15)

b = \frac{30x+150}{3x+15} ← factor numerator and denominator

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  = \frac{30}{3}

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7 0
3 years ago
Read 2 more answers
Find the distance between (-3, 11) and (-6,4).Round to the nearest then the
Likurg_2 [28]

Answer:

d =7.6

Step-by-step explanation:

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3 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
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