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Elis [28]
2 years ago
15

Find the degree of each polynomial and indicate whether the polynomial is a monomial, binomial or trinomial, or none of these. -

7^5abc
Mathematics
1 answer:
SIZIF [17.4K]2 years ago
7 0

Answer:

Monomial, degree is 5

Step-by-step explanation:

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(-3,4) m= -3 how would you solve this problem
nignag [31]

you would do PEMDAS

you do -3 times 4. then you do that and devide by -3.

5 0
3 years ago
Newton's law of cooling is:
Mnenie [13.5K]

Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

7 0
3 years ago
Ryan said that 212.314 73.243 285.557. Is Ryan's answer reasonable?
lord [1]
A.... maybe not sure
7 0
3 years ago
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

\frac{1}{2}=e^{.1x}

"Undo" that e by taking the ln of both sides:

ln(.5)=ln(e^{.1x})

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others.  Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

(2^2)^x-6(2)^x=-8

Now we will bring over the -8 by adding:

(2^2)^x-6(2)^x+8=0

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution.  Let's let u=2^x

When we do that, we can rewrite the polynomial as

u^2-6u+8=0

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor.  Now we have to put it back in:

2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

2^x=2^1

Now that the bases are the same, we can say that

x = 1

5 0
2 years ago
describe what would be true about the rate of change in a situation that could be represented by a graphed line and an equation
Andru [333]
If y=mx+b,

\frac { dy }{ dx } =m

You can't differentiate a constant, so dy/dx=m.
5 0
3 years ago
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