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Drupady [299]
3 years ago
13

What is the answer for this I need it asap

Mathematics
2 answers:
musickatia [10]3 years ago
7 0
$180 Earned and 3 weeks walking dogs
lawyer [7]3 years ago
5 0
You are starting at 0 dollars and it increases by $60 per week. The rate of chance is $60 per week.
You might be interested in
-6•(-18).......help this is due tomorrow
marin [14]
Two positives multiplied together give us a positive answer. and 18 • 6 is 108.
108 is the answer
hope this helped, xx
7 0
3 years ago
Read 2 more answers
If f(x) and f -1 (x) are inverse functions of each other and f(x) = 2x+5, what is f -1 (8)?
Aleksandr [31]

Answer: 3/2

Step-by-step explanation:

The first one: y=2x+5

X=2y+5.Then solve for y after switching x and y

X-5=2y, X-5/2=y

Since F-1= X-5/ 2, we plug the value of X in

8-5/2, so it should be 3/2

5 0
3 years ago
5. Find the general solution to y'''-y''+4y'-4y = 0
CaHeK987 [17]

For any equation,

a_ny^(n)+\dots+a_1y'+a_0y=0

assume solution of a form, e^{yt}

Which leads to,

(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0

Simplify to,

e^{yt}(y^3-y^2+4y-4)=0

Then find solutions,

\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}

For non repeated real root y, we have a form of,

y_1=c_1e^t

Following up,

For two non repeated complex roots y_2\neq y_3 where,

y_2=a+bi

and,

y_3=a-bi

the general solution has a form of,

y=e^{at}(c_2\cos(bt)+c_3\sin(bt))

Or in this case,

y=e^0(c_2\cos(2t)+c_3\sin(2t))

Now we just refine and get,

\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}

Hope this helps.

r3t40

5 0
3 years ago
Pls help asap i need help plz
shepuryov [24]

Answer:

False, It was till Pack 5 it was going up 3 every pack but at 5 it went up 13 cookies.

3 0
2 years ago
Read 2 more answers
ASAP someone please simply two exponents
zhannawk [14.2K]

Answer:

1.) m^{15}

2.) =\frac{1}{y^{15}}

Give me a comment if you want the explanation.

1.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

\left(m^3\right)^5=m^{3\cdot \:5}

\mathrm{Multiply\:the\:numbers:}\:3\cdot \:5=15

=m^{15}

2.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

\left(y^{-3}\right)^5=y^{-3\cdot \:5}

=y^{-3\cdot \:5}

=y^{-15}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\frac{1}{y^{15}}

4 0
3 years ago
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