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uranmaximum [27]
3 years ago
6

Please help! I will mark first answer brainliest and lots of points!!!! Jeremy is playing a game called “Rational Round Up” wher

e he has to collect all the numbers in a maze that are rational and then get to the end of the maze. When he collects a number he must prove it is rational by writing it as a quotient of two integers. Help him determine how to prove that each of the following numbers is rational. If you answer all I will appreciate that very much!
1. -18
2. 87.125
3. -30
4. -8.3
5. 58.25
6. 121
7. 4.5
Mathematics
2 answers:
natita [175]3 years ago
8 0
Shouldn’t it be 6. and 3. I hope it helped
Vladimir79 [104]3 years ago
7 0

Answer: All Options also - 9 + -9 = -18

Step-by-step explanation: The Reason It Is All Of Them Is Because A Rational Number Is A Fraction Integer or Some Kind of repeating decimal And They All Are In That Category.

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Scientists measure a bacteria population and find that it has 10,000 organisms. Five days later, they find that the population h
liraira [26]

Answer:

f(x)= 10,000(2)^x

Step-by-step explanation:

Exponential equations can be modeled by a(r)^x where a is the initial amount, r is the difference, and x is how many units the equation should be squared by.

7 0
3 years ago
What is the answer to (t+8)(-2)=12
Lunna [17]
(t + 8) (-2) = 12

Distributive property >>>      -2t + -16 = 12

Inverse operation >>>>    -2t + -16+16 = 12+16
 
Simplify >>>>                 -2t = 28

Divide >>>>                       -2t/-2 = 28/-2

Final answer >>>>              t = -14
8 0
3 years ago
Read 2 more answers
A ray that contains a point that is equidistant from both sides of the angle. *
seraphim [82]
<span>Consider a angle â BAC and the point D on its defector Assume that DB is perpendicular to AB and DC is perpendicular to AC. Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC Proof Consider triangles ΔADB and ΔADC Both are right angle, â ABD= â ACD=90 degree They have congruent acute angle â BAD and â CAD( since AD is angle bisector) They share hypotenuse AD therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
3 0
4 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
F m 1=7x and m 4=3x+20 what is the m 2
djyliett [7]
M2=5x+10
.................
4 0
3 years ago
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