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Sergio039 [100]
3 years ago
6

32 feet per sec squared into inches

Mathematics
2 answers:
jolli1 [7]3 years ago
6 0
Answer;
384 in/s2

Conversion rate:
1 ft/s2 = 12 in/s2

(a brainliest would be appreciated)
Natasha_Volkova [10]3 years ago
5 0

It would be 32 x 12 because there are 12 inches per feet. So the answer would be 384 inches per sec squared. If this is right could you possibly give me brainliest? Hope this helped.

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Answer:

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A parent made x cupcakes for each of the 109 students in the fourth grade. Which expression could be used to determine the total
Shkiper50 [21]

Answer:

C. 109x

Step-by-step explanation:

We know that there are 109 students in the fourth grade, who each have x cupcakes. In order to find the total amount of cupcakes, we could add all of the cupcakes together to get:

x + x + x + x + .... x

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Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)
zloy xaker [14]

Answer:

A)

k=0

B)

\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}

C)

\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}

Step-by-step explanation:

We are given the function:

\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}

A)

Given that h(1) = 20, we want to find <em>k</em>.

h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

\displaystyle (20) = 20e^{k(1)}

Simplify:

1= e^k

Anything raised to zero (except for zero) is one. Therefore:

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B)

Given that h(1) = 40, we want to find 2<em>k</em> + 1.

Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

\displaystyle (40) = 20e^{k(1)}

Simplify:

\displaystyle 2 = e^{k}

We can take the natural log of both sides:

\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}

By definition, ln(e) = 1. Hence:

\displaystyle k = \ln 2

Therefore:

2k+1 = 2\ln 2+ 1 \approx 2.3863

C)

Given that h(1) = 10, we want to find <em>k</em> - 3.

Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

\displaystyle (10) = 20e^{k(1)}

Simplfy:

\displaystyle \frac{1}{2} = e^k

Take the natural log of both sides:

\displaystyle \ln \frac{1}{2} = k\ln e

Therefore:

\displaystyle k = \ln \frac{1}{2}

Therefore:

\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931

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