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3 years ago

May you please help me??

Mathematics

2 answers:

TiliK225 [7]3 years ago

7 0

Answer: 14:00

Step-by-step explanation:

We have to add to the initial time the times that are consumed with each activity:

First time: 7:00

7hours 0 minutes

Three surgeries x two hours = 6 hours

7hours 0 minutes + 6 hours = 13hours 0 minutes

He drives 30 minutes

13 hours 0 minutes + 30 minutes = 13hours 30 minutes

He eats a snack 30 minutes

13hours 30 minutes + 30 minutes = 13 hours 60 minutes = 14hours 0 minutes

Answer: 14:00

<h2><em>Spymore</em></h2>

mart [117]3 years ago

3 0

Answer:

.....the answer is 1400

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2 years ago

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Answer. it fast please

mojhsa [17]

Answer:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

Step-by-step explanation:

Given

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Required

Solve:

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Simplify 6/10

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{9})^2$

Express 9 as 3^2

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{3^2})^2$

Apply law of indices

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{(3^2)^2}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{3^4}$

Express as a sing;e fraction

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3*5^2}{5^3*3^4}$

Apply law of indices:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^{3-2}*3^{4-3}}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^1*3^1}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5*3}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

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2 years ago

PLZ HELP!!! WELL WIN BRAINLIEST!!!! Describe an everyday situation that involves dividing a mixed number by a fraction.

Mekhanik [1.2K]

Answer:

Andy has to paint a 36 1/2 wall purple needs to hang a picture every 1/2 foot. How many paintings can he hang?

Step-by-step explanation:

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3 years ago

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Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

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3 years ago

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May you please help me??​

May you please help me??