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2 years ago

Which statement is true about the end behavior of the graphed function?

Mathematics

2 answers:

allochka39001 [22]2 years ago

8 0

Answer:

As the x-values go to positive infinity, the function's values go to negative infinity.

Step-by-step explanation:

Tems11 [23]2 years ago

6 0

Answer:

(SEE EXPLANATION FIRST)

Answer D is TRUE because on the graph if you were to keep going left even after the picture of the graph ends, the graph will still keep going up infinite

Step-by-step explanation:

If this is the graph you can read the answer

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HELP PLEASE this is due tomorrow for me!!

Debora [2.8K]

Answer:

YOUR ANSWER IS 9

Step-by-step explanation:

1.80/4= 0.45

2.70/6=0.45

5.40/12=0.45

0.45*20= 9$

5 0

3 years ago

A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t

Andreas93 [3]

Answer:

a) Side Parallel to the river: 200 ft

b) Each of the other sides: 400 ft

Step-by-step explanation:

Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be $F=L+2W$ .

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

$LW=80,000...(1)$

We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be $C=20L+5(2W)\Rightarrow 20L+10W$ .

From equation (1), we will get:

$L=\frac{80,000}{W}$

Upon substituting this value in cost equation, we will get:

$C=20(\frac{80,000}{W})+10W$

$C=\frac{1600,000}{W}+10W$

$C=1600,000W^{-1}+10W$

To minimize the cost, we need to find critical points of the the derivative of cost function as:

$C'=-1600,000W^{-2}+10$

$-1600,000W^{-2}+10=0$

$-1600,000W^{-2}=-10$

$-\frac{1600,000}{W^2}=-10$

$-10W^2=-1,600,000$

$W^2=160,000$

$\sqrt{W^2}=\pm\sqrt{160,000}$

$W=\pm 400$

Since width cannot be negative, therefore, the width of the fencing would be 400 feet.

Now, we will find the 2nd derivative as:

$C''=-2(-1600,000)W^{-3}$

$C''=3200,000W^{-3}$

$C''=\frac{3200,000}{W^3}$

Now, we will substitute $W=400$ in 2nd derivative as:

$C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05$

Since 2nd derivative is positive at $W=400$ , therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting $W=400$ in $L=\frac{80,000}{W}$ , we will get:

$L=\frac{80,000}{400}\\\\L=200$

Therefore, the side parallel to the river will be 200 feet.

4 0

3 years ago

Keith had 1 1/2 cups of flour.he used 5 2/3 cups to make bread.How many cups of flour does Keith have left

crimeas [40]

If Keith had 1 1/2 cups, he can't use <span>5 2/3 cups. He cant use more than what he had. There is error in this question.</span>

8 0

2 years ago

This box is a right rectangular prism with a base that has dimensions of 6 m and 7 m. The volume of the box is 378 m3. What is t

Over [174]

Answer:

29.07m³

Step-by-step explanation:

7+6=13

378÷13=29.07m³

5 0

3 years ago

Read 2 more answers

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

5 0

2 years ago