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Valentin [98]
2 years ago
8

PLEASE HELP!! WILL MARK BRAINIEST !!

Mathematics
1 answer:
SOVA2 [1]2 years ago
3 0

Answer:

(f+g)(2) = 10

Step-by-step explanation:

f(x)=2x^2+3x

g(x)=x-2

(f+g)(x) will be sum of f(x) and g(x)

(f+g)(x)  = 2x^2+3x + x - 2 = 2x^2+4x - 2

we have to find  (f+g)(2), for that we will put x = 2

(f+g)(x)  = 2x^2+4x - 2

(f+g)(2)  = 2*2^2+2*2 - 2\\(f+g)(2)  = 2*4+4 - 2\\(f+g)(2)  = 8+4 - 2 = 10

Thus,  (f+g)(2) is 10.

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5x maybe dont take my word

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24. If Jenna spent 30 hours on the computer this week. how much total time did she spend on
Verdich [7]

Option B) 3 hours is the time spent by Jenna on the e-mail this week.

<u>Step-by-step explanation:</u>

Jenna spent 30 hours on the computer this week.

The amount of time she spends on computer to work on various fields in given in the pie-chart.

This 30 hours represents her 100% work on computer.

From the figure shown,

She spent 10% of her total work on e-mail.

Therefore, 10% of 30 is the work done in hours by her on e-mail.

⇒ (10/100) × 30

⇒ 0.1 × 30

⇒ 3 hours

∴ Option B) 3 hours is the time spent by Jenna on the e-mail this week.

7 0
3 years ago
Find the area of this circle. Use 3 for a.A = 7r2=11 in[?] in?
liberstina [14]

SOLUTION

We want to find the area of the circle in the picture given.

We have been given the formula as

\begin{gathered} A=\pi r^2 \\ We\text{ are told to take }\pi\text{ as 3} \\  \end{gathered}

From the circle, the radius r = 11 in.

The area of the circle becomes

\begin{gathered} A=\pi r^2 \\ A=\pi\times r^2 \\ A=3\times11^2 \\ A=3\times11\times11 \\ A=363in^2 \end{gathered}

Hence the answer is 363 square-inches

7 0
8 months ago
Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
9 months ago
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