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Olin [163]
3 years ago
13

Determine if each pair of ratios or rates is equivalent. Explain your reasoning. 288 miles on 12 gallons of fuel; 240 miles 10 g

allons of fuel
Mathematics
1 answer:
lesantik [10]3 years ago
3 0
Yes they are equivalent. This is because if you simplify each fraction : 288/12 and 240/10 they both equal 24.
\frac{288}{12}  = 24
\frac{240}{10}  = 24
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16 2/3

you just need to multiply 4 1/6 by 4
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PLZ dont ignore me! I need help on this :(
uysha [10]
Answer: It’s the second one!

Explanation: The dot is open and pointing to the side of the numbers greater than -3. It has to be an open dot because it is “greater than -3” not “greater than or equal to-3”! :)
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Solve for angel GLN<br><br> please help asap, need help
likoan [24]

Answer:

119

Step-by-step explanation:

\mathrm{Supplementary\:angles\:add\:up\:to}\:180^{\circ }

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3 years ago
Maria rents a bicycle while on vacation. It costs a flat fee of $19.95 plus $5.50 an hour. Maria rents it for 4 hours. What is t
kolezko [41]

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7 0
3 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

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)

Replacing y with f⁻¹(x), we have

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:

brainly.com/question/9028678

5 0
3 years ago
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