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nordsb [41]
3 years ago
12

Is 9 greater then -8?

Mathematics
2 answers:
prohojiy [21]3 years ago
6 0

Answer:

yes, anything negative is automatically less than anything positive...

Alex Ar [27]3 years ago
4 0

Answer:

Yes. Bc 8 have a negative sign infront of it so 9 greater

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Use the long division method to divide 120÷24​
Natali5045456 [20]

Step-by-step explanation:

120÷24

    

    0   24⟌120

    0   24⟌120   0  

    0   24⟌120   -0      1

    00  24⟌120   -0      12 

   0   24⟌120   -0      12 

    00  24⟌120   -0      12    - 0     12 

   00  24⟌120   -0      12    - 0     120

4 0
3 years ago
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30 POINTS!! “Use the figures to help you complete the statements”<br><br> Help pls
ELEN [110]

Answer:

<u>6</u>

<u>4</u>

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5 0
2 years ago
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What is the output of the function f(p) = 3p – 2 when the input is 2?
vovikov84 [41]
Just plug in 2 for p. 3*2 - 2 = 6-2 = 4
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3 years ago
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22 gallons of gas cost $65.56. What is the unit rate for 1 gallon of gas?
Vika [28.1K]
It’s 2.98 explanation:you divide $65.56 and 22.
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The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

Substituting into the ODE, we get

v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0
v''x\ln x+(2+\ln x)v'=0

Setting w=v', we end up with the linear ODE

w'x\ln x+(2+\ln x)w=0

Multiplying both sides by \ln x, we have

w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x

we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
w=\dfrac{C_1}{x(\ln x)^2}

Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
v=-\dfrac{C_1}{\ln x}+C_2

So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
4 0
3 years ago
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