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3 years ago

5

Please help me babes!!

2 answers:

musickatia [10]3 years ago

8 0

C it is indeed cba - efd

abruzzese [7]3 years ago

6 0

The correct answer would be C) CBA-EFD

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At the corner store near Billy’s

nikklg [1K]

Answer:

d = 0.0175n ---> required equation

Billy can buy 285.714 gram nut with $5

Step-by-step explanation:

cost of 100g loose nut = $1.75

dividing LHS and RHS by 100

cost of 100/100g loose nut = $1.75

Thus, cost of 1 gm loose nut = $0.0175

let the weight of loose nut be n gm

Multiplying LHS and RHS by n

cost of x g loose nut = $0.0175*n = $0.0175n

It is given that Billy spent d dollars to buy n gm nuts

thus,

d = 0.0175n ---> required equation

________________________________________________

He spent $5 to buy nuts

substituting value of d as 5 we have

0.0175n = 5

=>n = 5/0.0175 = 285.714

Thus, Billy can buy 285.714 gram nut with $5.

7 0

3 years ago

I do not know how to do this need help

Diano4ka-milaya [45]

Remark

First of all you have to declare the meaning of g(f(x)) After you have done that, you have to make the correct substitution.

Givens

f(x) = 4x^2 + x + 1

g(x) = x^2 - 2

Discussion

What the given condition g(f(x)) means is that you begin with g(x). Write down g(x) = x^2 - 2

Wherever you see an x on either the left or right side of the equation, you put fix)

Then wherever you see f(x) on the right you put in what f(x) is equal to.

Solution

g(x) = x^2 - 2

g(f(x)) = (f(x))^2 - 2

g(f(x)) = [4x^2 + x + 1]^2 - 2

f(x)^2 =

4x^2 + x + 1

<u>4x^2 + x + 1</u>

16x^4 + 4x^3 + 4x^2

4x^3 + x^2 + x

<u> 4x^2 + x + 1</u>

16x^4 + 8x^3 + 9x^2 + 2x + 1

Answer

g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x + 1 - 2

g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x - 1

5 0

4 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

Lines g,h, and i are all parallel and m------->1 =55*

Hunter-Best [27]

It would be C:55,
hope this helps u out

7 0

3 years ago

Chase puts 2/7 of a cup of trail mix into 3/4 of a cup size bags. How many bags can he make?

wolverine [178]

Answer:

i have not done this in a year or 2 but i believe it is

2 5/28

Step-by-step explanation:

i may be wrong

7 0

3 years ago

Read 2 more answers