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gtnhenbr [62]
3 years ago
6

Number 13 and 14 don’t matter but help ‼️

Mathematics
1 answer:
DIA [1.3K]3 years ago
6 0
Find a common denominator for every single pair of fractions
that will help you find the answer
i can do a couple but not all, hopefully my explanation will help you do the rest

3/5 and 3/3
3/15 and 3/15
9/15 and 15/15
15/15 is greater than 9/15
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Step-by-step explanation:

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Isaac surveyed the employees at a law firm. Each employee was asked to record their highest level of education completed. The re
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26

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I need to know the answer to p=4x-140
dalvyx [7]
This is not look like something that could be answered, check around the paper for a graphing box.
This looks like a formula for a line on a graph: y=Mx+b. I can help you if this is the case, 1. Put a point at -140 on the Y axis (up and down) 2. Move up 1 and over 4and put a dot there( you could multiply the 1 and 4 to cover a larger area) because it goes all the way to -140.
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3 years ago
25. Several functions represent different savings account plans. Which functions are linear?
Mice21 [21]

Answer:

b, c, d

Step-by-step explanation:

y=kx+b

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y=kx

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y=0,5x-right

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8 0
3 years ago
Find the general solution of the given differential equation.
Elina [12.6K]

Answer:

y=-x\,cos\,x+Cx

There are no transient terms.

Step-by-step explanation:

Given: x\,\frac{dy}{dx} -y=x^2\,sin\,x

To find: general solution of the differential equation and the transient terms in the general solution.

Solution:

For an equation of the form \frac{dy}{dx}+yp(x)=q(x),

solution is given by ye^{\int {p(x)} \, dx } = ∫ q(x)e^{\int {p(x)} \, dx } dx

The given equation x\,\frac{dy}{dx} -y=x^2\,sin\,x can be written as \frac{dy}{dx}-\frac{y}{x}=x\,sin\,x

Here,

p(x)=\frac{-1}{x}\,,\,q(x)=x\,sin\,x

e^{\int{p(x)} \, dx } =e^{\int{\frac{-1}{x} } \, dx } =e^{-ln(x)} =e^{ln(x^{-1} )}=x^{-1}=\frac{1}{x}

So,

the solution is \frac{y}{x}=\int \frac{1}{x}x\,sin\,x\,dx

\frac{y}{x} =\int\,sin\,x\,dx\\\\\frac{y}{x} =-cos\,x+C\\y=-x\,cos\,x+Cx

Here, C is a constant.

Transient term is a term such that it tends to 0 as x → ∞

Here, there does not exist any term that tends to 0 as x → ∞

So, there are no transient terms.

7 0
3 years ago
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