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Tresset [83]
3 years ago
15

For every hour that passes, the temperature drops by 6. What is the temperature change after 4 hours

Mathematics
2 answers:
Ber [7]3 years ago
8 0

Answer:

24

Step-by-step explanation:

1 hour = 6

So, 6 times 4 equals 24 which is how much temperature that changed.

choli [55]3 years ago
7 0

Answer: 24

Step-by-step explanation:

1 hour = 6

So, 6 times 4 equals 24 which is how much temperature that changed.

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What is the unit rate for meters per second if a car travels 374 meters in 17 seconds?
Vikentia [17]

Answer:

22 meters per second.

Step-by-step explanation:

374 divided by 17 = 22.

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HELP IM IN A QUIZZ... In a certain game, players earn 10 points for each question answered correctly and -5 points for questions
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Answer:

-10

Step-by-step explanation:

Just set up a simple equation.

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A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic
Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

\mathbf{Bars = 6}

\mathbf{Chocolate = 3}

\mathbf{Custard= 3}

The total number of donuts in the box is:

\mathbf{Total= 6 + 3 + 3}

\mathbf{Total= 12}

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}

So, we have:

\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}

Simplify

\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}

Multiply

\mathbf{Pr = \frac{9}{132}}

Divide

\mathbf{Pr = 0.068}

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068

Read more about probabilities at:

brainly.com/question/9000575

7 0
2 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

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