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3 years ago

- Functions F and A are defined by these equations below. Which function has a greater value when x is 2.5?

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

3 0

Answer:

The answer is D- A(x)=25+10x

Step-by-step explanation:

I cant really explain how i got the answer but i got it right

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Pls answer I will give brainliest

Levart [38]

Answer: 108 ft³

Step-by-step explanation:

V= 4/3πr³

Facts:

Pi= 3

Diameter = 6

Step1; find the radius

Diameter is the length acaross the whole circle, but the radius is just 1/2 the circle. Therefore, if the diameter is 6, then the radius can be found by finding 1/2 of 6.

1/2 of 6= 1/2x6= 3.

R= 3

Step 2: Plug in the numbers (into the formula)

PEMDAS

(exponents go first)

(then multiply from left to right)

V= 4/3πr³

V= 4/3x3x3³

V= 4/3x3x (3x3x3)

V= 4/3x3x27

V= (4/3x3) x 27

V= 12/3x27

V= 4x27

V= 108

Volume= 108 ft³

Hope this helps :)) Also brainliest would be appreciated, i worked pretty hard on this thanks! :)

7 0

2 years ago

Read 2 more answers

write an equation that demonstrates the relationship between x and y for the points plotted on the coordinate grid

svetoff [14.1K]

Answer: y=3x-7

Explanation:

The y-intercept is -7 and the line is going up by a rate of 3.

5 0

3 years ago

Solve for AB a: 3.6 b: 15 c: 4 d: 10

IrinaK [193]

A because 9 divided by 15 equals 0.6. Then I multiply 0.6 times 6 which equals 3.6. I didn’t 9 divided by 15 because they are congruent.

4 0

3 years ago

Evaluate square root of 16 minus x end root when x = 8.

My name is Ann [436]

$\sqrt{16-x}$
x=8
16-8=8
$\sqrt{8}$

Use a calculator to find the square root of 8.

7 0

3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$