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2 years ago

Deshaun started baking at 9:25 AM and finished at 10:41 AM How long did it take him? Give your answer in hours and minutes

Mathematics

1 answer:

Zepler [3.9K]2 years ago

7 0

Answer:

1 hour and 16 minutes

Step-by-step explanation:

you just have to count up from 9:25 to 10:41 and translate the amount of minutes you counted into hours

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-5w = -24.5 what is it.

pav-90 [236]

Answer:

w=4.9

Step-by-step explanation:

If you are solving for w, you want to isolate the variable. If it is easier to write it out, it would look something like this..

-5 x w = -24.5

To get w by itself, use the inverse of multiplication, which is division. Divide the expression to the left of the equal sign by -5. Remember, what you do to one side, you must do to the other. When you divide -24.5 by -5, you end up with

w=4.9

6 0

3 years ago

3(x+4)=2x+6 can you help me solve this

Vladimir [108]

Answer:

x=-6

Step-by-step explanation:

3(x+4)=2x+6

3x+12=2x+6

x+12=6

x=-6

5 0

3 years ago

Percy paid $24.10 for a basketball the price of the basketball was $22.99 what was the sales tax rate

Nat2105 [25]

To find the sales tax rate, you need to find what percentage of the base price the difference in cost is.

First, subtract 22.99 from 24.10:

24.1-22.99=1.11

Now, divide 1.11 by 22.99:

1.11/22.99=0.0483

Multiply 0.0483 by 100 to convert the decimal to a percentage:

0.0483*100=4.83

The sales tax rate was 4.83%.

Hope this helps!!

3 0

3 years ago

4(2v-6)-12v<br> Topic Name<br> Using distribution and combining like terms to simplify: Univariate

iragen [17]

Answer:

-24+4v

Step-by-step explanation:

8v-24-12v

-24+4v

3 0

3 years ago

Evaluate using L'Hopital's rule:<br><br> <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%201%5E%7B%2B%7D%7D%20%28x%20-%201

SSSSS [86.1K]

Answer:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

Step-by-step explanation:

We are given:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)$

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

$=(1-1)^\ln(1)}=0^0$

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

$y=(x-1)^\ln(x)$

By taking the natural log of both sides:

$\ln(y)=\ln(x)\ln(x-1)$

And by taking the limit as x approaches 1 from the right of both functions:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)$

Rewrite:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}$

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}$

Now, by using direct substitution, we will acquire:

$\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}$

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))$

Finally, direct substitution yields:

$\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0$

Thus:

$\displaystyle \lim_{x\to 1^+}\ln(y)=0$

By the Composite Function Property for limits:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0$

Raising both sides to e produces:

$\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0$

Therefore:

$\displaystyle \lim_{x\to 1^+}y=1$

Substitution:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

4 0

2 years ago

Deshaun started baking at 9:25 AM and finished at 10:41 AM How long did it take him? Give your answer in hours and minutes​

Deshaun started baking at 9:25 AM and finished at 10:41 AM How long did it take him? Give your answer in hours and minutes