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3 years ago

Use the function below to find F(4).

2 answers:

3 0

To find F(4), you simply plug 4 in for x. That means multiplying 5 by 1/2x to the 4th power. That comes out to .3125, or D. (5/16)

serious [3.7K]3 years ago

3 0

Answer:

The answer is the option D

$(\frac{5}{16})$

Step-by-step explanation:

we have

$f(x)=5(\frac{1}{2})^{x}$

we know that

To find $f(4)$

substitute the value of $x=4$ in the function

$f(4)=5(\frac{1}{2})^{4}$

$f(4)=5(\frac{1}{16})$

$f(4)=(\frac{5}{16})$

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An orchestra has 100 musicians. 40/100 of them play string instruments- violin,viola,

IrinaK [193]

That's 40%

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6 0

3 years ago

Read 2 more answers

Through (-5.-2) with a slope = 4/5

eduard

It should be -3 I hope this helps

7 0

3 years ago

The average of 100 measurements is 23 and the average of 50 additional measurements is 27.

Elden [556K]

Total value of 100 measurements
100 * 23
2300

Total value of additional 50 measurements
50 * 27
1350

Total value of 150 measurements
2300 + 1350
3650

Average of 150 measurements
3650 / 150
24.3

25 > 24.3

Answer is B

8 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$