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Tema [17]
4 years ago
8

What is the answer to this problem please -5a +3(3a -9)=

Mathematics
1 answer:
Basile [38]4 years ago
8 0
You can use the distributive property
-5a + 3 (3a-9)
-5a + 9a -27
4a - 27 = ?
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Step-by-step explanation:

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Can someone help me solve this problem (graphing/geometry)
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Step-by-step explanation:

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What's the zero product property for (x+13)(x-7)=0
umka21 [38]
(x+13)*(x-7)=0
Split into possible cases :
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x+13=0
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3 years ago
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Estimate the square root of 250 to two decimal places.
Andre45 [30]

Answer:

15.81

Step-by-step explanation:

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3 0
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

6 0
4 years ago
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