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ivanzaharov [21]
4 years ago
8

Solve for x: 3(x + 1) = −2(x − 1) + 6 A. 1 B. 4 C. 5 D. 25

Mathematics
2 answers:
olga_2 [115]4 years ago
8 0
<span>3(x+1)=-2(x-1)+6
3x+3=-2x+8
5x=5
x=1

Thus, A would be the answer.</span>
zysi [14]4 years ago
6 0

Answer:

Correct option is:

A. 1

Step-by-step explanation:

3(x + 1) = −2(x − 1) + 6

⇒ 3x+3= -2x+2+6

⇒ 3x+3= -2x+8

⇒ 3x+2x=8-3

⇒ 5x=5

dividing both sides by 5

⇒ x=1

Hence, the correct option is:

A. 1

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katen-ka-za [31]

Answer:

  33, 39, 45, 51

Step-by-step explanation:

The general term of an arithmetic sequence is given by the formula ...

  an = a1 + d(n-1) . . . where a1 is the first term and d is the common difference

Comparing this formula to the one you are given, you see that ...

  a1 = 33, d = 6

This means the first term is 33, and each successive term is 6 more than the previous one. The first 4 terms are ...

  33, 39, 45, 51

6 0
3 years ago
Please help ASAP<br><br> If f(x) = x^2-6x+2 and g(x) =sqrt x find a.) (fog) (x)<br> b.) (gof) (-2)
Rudik [331]
\large\begin{array}{l} \left\{\!\begin{array}{l} \mathsf{f(x)=x^2-6x+2}\\ \mathsf{g(x)=\sqrt{x}}\\ \end{array}\right. \end{array}


\large\begin{array}{l} \textsf{a) }\mathsf{(f\circ g)(x)}\\\\ =\mathsf{f\big[g(x)\big]}\\\\ =\mathsf{\big[g(x)\big]^2-6\cdot g(x)+2}\\\\ =\mathsf{\big[\sqrt{x}\big]^2-6\sqrt{x}+2}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(f\circ g)(x)=x-6\sqrt{x}+2} \end{array}}\qquad\checkmark \end{array}

______


\large\begin{array}{l} \textsf{b) }\mathsf{(g\circ f)(-2)}\\\\ =\mathsf{g\big[f(-2)\big]}\\\\ =\mathsf{\sqrt{f(-2)}}\\\\ =\mathsf{\sqrt{(-2)^2-6\cdot (-2)+2}}\\\\ =\mathsf{\sqrt{4+12+2}}\\\\ =\mathsf{\sqrt{18}}\\\\ =\mathsf{\sqrt{3^2\cdot 2}}\\\\ =\mathsf{3\sqrt{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{(g\circ f)(-2)=3\sqrt{2}} \end{array}}\qquad\checkmark \end{array}

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If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2181559


\large\textsf{I hope this helps. :-)}


Tags: <em>composite function composition evaluate algebra</em>

7 0
3 years ago
In 2012 your car was worth $10,000. In 2014 your car was worth $8,850. Suppose the value of your car decreased at a constant rat
monitta

Answer:

The function to determine the value of your car (in dollars) in terms of the number of years t since 2012 is:

f(t) = 10000(0.9407)^t

Step-by-step explanation:

Value of the car:

Constant rate of change, so the value of the car in t years after 2012 is given by:

f(t) = f(0)(1-r)^t

In which f(0) is the initial value and r is the decay rate, as a decimal.

In 2012 your car was worth $10,000.

This means that f(0) = 10000, thus:

f(t) = 10000(1-r)^t

2014 your car was worth $8,850.

2014 - 2012 = 2, so:

f(2) = 8850

We use this to find 1 - r.

f(t) = 10000(1-r)^t

8850 = 10000(1-r)^2

(1-r)^2 = \frac{8850}{10000}

(1-r)^2 = 0.885

\sqrt{(1-r)^2} = \sqrt{0.885}

1 - r = 0.9407

Thus

f(t) = 10000(1-r)^t

f(t) = 10000(0.9407)^t

7 0
3 years ago
Over the past 7 years, Ms. McDonald raised corn for her pigs. She has found that 5 out of 7 seeds produce corn. She wants to hav
Eva8 [605]

Answer: 14,000 seeds

Step-by-step explanation: hope this helps<3

7 0
3 years ago
What is tan theta when csc theta =2 sqrt 3?
Slav-nsk [51]

Answer:

\displaystyle \tan\theta=\frac{\sqrt{11}}{11}

Step-by-step explanation:

<u>Trigonometric Identities</u>

If the trigonometric function value of an angle is given, we can find the rest of the trigonometric values of the angle by using one or more of the fundamental identities.

To solve this problem, we need to use the following identities:

\displaystyle \cot^2\theta=\csc^2\theta-1

\displaystyle \tan\theta=\frac{1}{\cot\theta}

Since we are given:

\displaystyle \csc\theta=2\sqrt{3}

And the angle is in the first quadrant, calculate the cotangent:

\displaystyle \cot^2\theta=(2\sqrt{3})^2-1

\displaystyle \cot^2\theta=4*3-1=11

\displaystyle \cot\theta=\sqrt{11}

The tangent is:

\displaystyle \tan\theta=\frac{1}{\sqrt{11}}

Rationalizing the denominator:

\displaystyle \tan\theta=\frac{1}{\sqrt{11}}*\frac{\sqrt{11}}{\sqrt{11}}

\boxed{\displaystyle \tan\theta=\frac{\sqrt{11}}{11}}

7 0
3 years ago
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