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atroni [7]
3 years ago
15

fire work a travels at the speed 300 ft/s firework b travels 240 ft/s. firework b is launched 0.25s before firework a. how many

seconds after firework b launches will both fireworks explode?
Mathematics
1 answer:
storchak [24]3 years ago
7 0

Answer:

Both fireworks  will explode after 1 seconds after firework b launches.

Step-by-step explanation:

Given:

Speed of fire work A= 300 ft/s

Speed of Firework B=240 ft/s

Time before which fire work b is launched =0.25s

To Find:

How many seconds after firework b launches will both fireworks explode=?

Solution:

Let t be the time(seconds) after which both the fireworks explode.

By the time the firework a has been launched, Firework B has been launch 0.25 s, So we can treat them as two separate equation

Firework A= 330(t)

Firework B=240(t)+240(0.25)

Since we need to know the same time after which they explode, we can equate both the equations

330(t) = 240(t)+240(0.25)

300(t)= 240(t)+60

300(t)-240(t)= 60

60(t)=60

t=\frac{60}{60}

t=1

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Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
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a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

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So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

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