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Trava [24]
3 years ago
11

90 PTS! Solve the system of equations below by graphing.

Mathematics
2 answers:
Mice21 [21]3 years ago
5 0

Answer:

1. x=  0.56+2

2. y=x2−3x−1

Step-by-step explanation:

geniusboy [140]3 years ago
3 0

Answer:

1. x=  0.56+2

2. y=x2−3x−1

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Explain how you can use a model 4 x 1/3
Romashka-Z-Leto [24]

You can use it to find the area of something



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3 years ago
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What’s the square root of 1700
leva [86]

Answer:

41.2310562562

Step-by-step explanation:

by use of calculator

8 0
3 years ago
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What are the terms in the expression 2x−4y+8 ?
vazorg [7]

Answer:

A. 2x, -4y, and 8

Step-by-step explanation:

If there is a minus sign in front of the 4 (-), add that to the expression. What you are just simply doing is separating the numbers up. For example:

12x - 8y + 4x

Now here, you have two of the same variables (x, y, etc). So, what you do is look at the last number that has the same variables, which is 4, and look at what the problem you will be solving, which is addition. So, very simply, you add then together!

12× + 4× = 16×

As you can see I kept the same variable. This is because, well, it is the same! Simply, just substitute in the 16× with the 8y. Now here is the tricky part, for some people. Do you see that there is a negative sign in front of the 8 (-)? Well! You have to substitute that in with the expression. No adding this or anything, just simply slide it next to the 16× because, we can not add nor subtract it with the 8y just because it has a different variable.

Your example answer would be: 16× - 8y

Hope this helps!

P.S. if you think this helped you at all, Brainliest me if ya want to. Have a great day!

5 0
2 years ago
Use the image above to write a conjecture about regular polygons and lines of symmetry
Tatiana [17]
To find conjecture about regular polygon, use the formula \frac{360}{n}

Where, n = number of sides of the regular polygone

1) Triangle
n =3
So, Conjecture of the triangle = \frac{360}{3} = 120

2) Square
n =4
So, Conjecture of the square = \frac{360}{4} = 90

3) Pentagone
n = 5
So, Conjecture of the pentagon = \frac{360}{5} = 72

4) Hexagone
n = 6
So, Conjecture of the pentagon = \frac{360}{6} = 60.
7 0
3 years ago
Read 2 more answers
(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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