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3 years ago

12

HELP IM SO SORRY IJAGEHQKSHAQHGSVWHSBWBSNSJWBSBEJEJEJEJJEEJ JUST HELP. JUST HELP ME OGMFBSBSBGSGSBSBANSJSJSJS NUMBERS 13 and 14!

?!!,!!!!!!!!!!!!!!!!!!

1 answer:

xxMikexx [17]3 years ago

4 0

14.The answer is 10 Celsius

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tom’s first carry of a football game resulted in a 5 yard loss his next carry was 12 1/2 if he finished the game with 3 carries

Mnenie [13.5K]

Answer:

18.5 yards

Step-by-step explanation:

Tom's first carry was -5 yards and his second was +12.5 yards. If we know he finished with 26 yards we can make a simple equation to try and find the distance of his third carry. We let the letter $c$ be a variable that represents the distance of his third carry.

$(-5) + (12.5) + c = 26$

This means his first carry + second carry + third carry = carry total

We can simply the left hand side

$7.5 + c = 26$

Now we move the 7.5 to the right side by subtracting it.

$c = 26 - 7.5 \\c = 18.5$

So Tom's third carry was for 18.5 yards!

Hope this helps!

5 0

3 years ago

Standard form for 1x1+9x1/10+8x1/100+1x1/1000

raketka [301]

It equals because when you multiply it all up

7 0

3 years ago

Read 2 more answers

Write the exponent for 4^? = 64 What number is the question mark? *

Elden [556K]

The ? Is 3

As 4^3 = 64

4x4x4= 64

5 0

3 years ago

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:

$k=0$

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:

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

3 0

2 years ago

Mary ran 2 miles in about 23 minutes. If she continued at the same pace, how long will it take her to run 10 miles?

8090 [49]

Answer:

115 minutes ( 1 hour and 55 minutes)

Step-by-step explanation:

Divide 23 by 2 to find out how fast she can run 1 mile

11.5

Now multiply 11.5 by 10

115 minutes

Hope this helped :)

Please consider Brainliest :)

7 0

3 years ago