.......... help please :)

If f(x)=x+3 and g(x)= x^2 -4 find the expression for g(f(x))

Maksim231197 [3]

Answer:g(f(x))= x^2 +6x+5

Step-by-step explanation:

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

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

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

g(f(x))= x^2 +6x+9-4

g(f(x))= x^2 +6x+5

4 0

3 years ago

5/8 of the cats perferred drinking with there tongue but 3/16 wanted a straw if there was 39 cats that gave one of these 2 answe

AveGali [126]

Answer:

208cats

Step-by-step explanation:

⅝=tongue

³/16=straw=39cats

⅝+³/16=13/16

meaning total number of cats is 16/16

If ³/16=39cats

16/16=?

39×16/3

=208cats

6 0

3 years ago

Two trains are moving towards each other. One train moves at a speed of 50 mph, and the other train travels at 60 mph. How soon

Margaret [11]

I think that they will meet in 3 hours and 11 min but I’m not too sure

4 0

3 years ago

Read 2 more answers

jenna's rectangular garden borders a wall. she buys 80 ft of fencing. what are the dimensions of the garden that will maximize i

FrozenT [24]

Answer:

The dimensions are x =20 and y=20 of the garden that will maximize its area is 400

Step-by-step explanation:

Step 1:-

let 'x' be the length and the 'y' be the width of the rectangle

given Jenna's buys 80ft of fencing of rectangle so the perimeter of the rectangle is 2(x +y) = 80

x + y =40

y = 40 -x

now the area of the rectangle A = length X width

A = x y

substitute 'y' value in above A = x (40 - x)

A = 40 x - x^2 .....(1)

Step :2

now differentiating equation (1) with respective to 'x'

$\frac{dA}{dx} = 40 -2x$ ........(2)

Find the dimensions

$\frac{dA}{dx} = 0$

40 - 2x =0

40 = 2x

x = 20

and y = 40 - x = 40 -20 =20

The dimensions are x =20 and y=20

length = 20 and breadth = 20

Step 3:-

we have to find maximum area

Again differentiating equation (2) with respective to 'x' we get

$\frac{d^2A}{dx^2} = -2$

Now the maximum area A = x y at x =20 and y=20

A = 20 X 20 = 400

Conclusion:-

The dimensions are x =20 and y=20 of the garden that will maximize its area is 400

verification:-

The perimeter = 2(x +y) =80

2(20 +20) =80

2(40) =80

80 =80

8 0

3 years ago

A tunnel is built in form of a parabola. The width at the base of tunnel is 7 m. On

uysha [10]

Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

$h(x)=ax(x-7)$ ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

$4=a(3)((3)-7)$

$4=a(3)(-4)$

$\dfrac{4}{(3)(-4)}=a$

$-\dfrac{1}{3}=a$

Putting $a=-\dfrac{1}{3}$ , we get

$h(x)=-\dfrac{1}{3}x(x-7)$ ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

$h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)$

$h(1.5)=-(0.5)(-5.5)$

$h(1.5)=2.75$

Since h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

6 0

3 years ago