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4 years ago

5

Please help? It’s a basic time question and I don’t get it. Either I’m really dumb or it’s to complicated

2 answers:

Y_Kistochka [10]4 years ago

5 0

Answer: D 1 hour 56 minutes

allsm [11]4 years ago

3 0

Answer:

D. 1 hour 56 minutes

Step-by-step explanation:

Well, if you add 2 hours to 12:14, it would be 2:14, so you know it's less than 2 hours. B and D are the only choices that have 1 hour. So, if Aaron left at 2:10, you have to subtract 4 minutes from 2 hours. That would be 1 hour and 56 minutes Aaron spent at the mall.

I hope you understood!! (It's ok if you didn't get it...some things can seem complicated at first!) Have an amazing day C:

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kogti [31]

v=d/t

44=d/5

d=220 mile

v=220/11

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3 years ago

Pleaseee help me I’m stuck

antiseptic1488 [7]

Answer:

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Step-by-step explanation:

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3 years ago

What does the pH of a solution represent?

lana [24]

Answer:

D. The pH indicates how acidic or basic a solution is.

Step-by-step explanation:

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7 0

3 years ago

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- $x^3$

6 0

4 years ago

Read 2 more answers

one of the solution to this problem below is 8. what is the other solution? C^2-64=0 ( please show your work )

iris [78.8K]

${c}^{2} - 64 = 0$
This can also be represented as:
${c}^{2} - {8}^{2} = 0$
Using difference of two squares;
$(c - 8)(c + 8) = 0$
$c - 8 = 0 \: or \: c + 8 = 0 \\ \\ c = 8 \: or \: c = - 8$

Hope I helped? ☺️

5 0

3 years ago