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kotykmax [81]
3 years ago
11

Compare the functions shown below: f(x) cosine graph with points at 0, negative 1 and pi over 2, 1 and pi, 3 and 3 pi over 2, 1

and 2 pi, negative 1 g(x) x y −6 −11 −5 −6 −4 −3 −3 −2 −2 −3 −1 −6 0 −11 h(x) = 2 cos x + 1 Which function has the greatest maximum y-value?

Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

  f(x) and h(x) have the same maximum value: 3

Step-by-step explanation:

The maximum value of f(x) is 3 at (π, 3).

The maximum value of g(x) is -2 at (-3, -2).

The maximum value of h(x) is 3 at (0, 3).

Both f(x) and h(x) have the same (greatest) maximum value.

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What is the value of b<br> 2b + 8 − 5b + 3 = −13 + 8b − 5
Yuri [45]

Answer:

See below.

Step-by-step explanation:

2b + 8 − 5b + 3 = −13 + 8b − 5

Reorder like terms.

2b-5b+8+3=8b-13-5

Combine those like terms. Then solve.

-3b+11=8b-18

      -11       -11

-3b=8b-29

-8b -8b

-11b=-29

/-11    /-11

b=2.64

-hope it helps

3 0
2 years ago
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Law Incorporation [45]

Answer

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3 years ago
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You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along
sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
3 years ago
If f(x) = 2x – 6 and g(x) = x – 2x2, find each value.<br><br><br> f(h + 9)
Zigmanuir [339]

Answer:

Solution for If f(x) = 2x – 6 and g(x) = x – 2x2, find each value. f(2)

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Answer:

A.) 40.506 cm is the answer

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