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Nataly [62]
2 years ago
10

Show work for answer please​

Mathematics
1 answer:
Ghella [55]2 years ago
8 0

Answer:

x = - 3

Step-by-step explanation:

Given

7 + x - 3 = x - 5 - 3x ( simplify both sides )

4 + x = - 2x - 5 ( add 2x to both sides )

4 + 3x = - 5 ( subtract 4 from both sides )

3x = - 9 ( divide both sides by 3 )

x = - 3

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Billy wants to make trail mix using almonds, peanuts, and raisins. He wants to mix one part almonds, two parts peanuts, and thre
Y_Kistochka [10]
X = price of the ingredients

so based on the proportion , we can write that :
Almonds needed = x     =                    $ 12 x
peanut needed   =  2x   = 2x * $ 8     =  $ 16x
raisins needed     = 3x  =  3x * $6      =  $ 18x

Equation = 
12x + 16x + 18x  ≤  $ 23

46x                  =     $23

x    =  $ 0.5

The pound of mix he can made :

0.5 * 1 pound of almond  + 0.5 x  * 2 pound of peanuts + 0.5 * 3 pound of raisins

= 3 Pound of mix

4 0
3 years ago
What is the slope of the line graphed below
Solnce55 [7]
Slope is 1/3

Equation of the line is:

y = 1/3x - 8
6 0
3 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
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gavmur [86]

Answer:

XYZ is similar to ABC

Step-by-step explanation:

take a ruler and measure the sides

6 0
3 years ago
What is The simple interest earned on a deposit of $3000 at 6% for five years
Norma-Jean [14]

Answer: $900

Step-by-step explanation:

The simple interest is calculated using the formula:

(P × R × T)/100

where,

P = Principal = $3000

R = Rate = 6%

T = Time = 5 years

Simple Interest = (P × R × T)/100

= ($3000 × 6 × 5)/100

= 90000/100

= $900

Therefore, the simple interest is $900

4 0
2 years ago
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