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Gnesinka [82]
3 years ago
5

DOSE EDENUITY SUCK? I WANA KNOW!

Mathematics
2 answers:
allochka39001 [22]3 years ago
8 0

Answer:

not really I think they just need an update to make it a little more better

Step-by-step explanation:

nadya68 [22]3 years ago
6 0

Answer:

If it involves math then yea

Step-by-step explanation:

trust me bro

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Find the measure of x and y.
cupoosta [38]

Create a proportion for x first. Use known values.

\frac{4}{x} =\frac{12}{21}

84 = 12x

Divide both sides by 12.

x = 7

---

Create a proportion for y.

\frac{12}{y} =\frac{4}{10}

120 = 4y

Divide both sides by 4.

y = 30

\boxed {x = 7,~y = 30}

7 0
3 years ago
If g(x) = 8x^6 and f(x) = log4 (2x) then f(g(x)) = ?
Lena [83]

\begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\begin{cases} g(x)=8x^6\\\\ f(x)=\log_4(2x) \end{cases}\qquad \qquad \begin{array}{llll} f(~~g(x)~~)=\log_4[~2g(x)~] \\\\\\ f(~~g(x)~~)=\log_4[~2(8x^6)~] \end{array} \\\\\\ f(~~g(x)~~)=\log_4(16x^6)\implies f(~~g(x)~~)=\log_4(4^2x^6) \\\\\\ f(~~g(x)~~)=\log_4(4^2)~~ + ~~\log_4(x^6)\implies \boxed{f(~~g(x)~~)=2~~ + ~~6\log_4(x)}

8 0
3 years ago
Write two inequalities that compare 0 and -8
Mrrafil [7]

Answer:

????????

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What whole number when rounding to the nearest hundred is 700
arlik [135]

Answer:

honestly quite a lot

Step-by-step explanation:

any number from 650-699, and also from 700-749

8 0
3 years ago
Read 2 more answers
Consider a population of rabbits in a region with unlimited food resources. Suppose that the growth rate is proportional to the
Ivahew [28]

Answer:

1) y(t) = \frac{Ce^{4t} +3}{4}

2) 5 = \frac{C e^0 +3}{4}

20 = C +3

C= 20-3=17

So then the model would be given by:

y(t) = \frac{20e^{4t} +3}{4}

Step-by-step explanation:

For this case we have the following differential equation:

y' = \frac{dy}{dt} = 4y-3

We can do this using algebra:

\frac{dy}{4y-3} = dt

Part 1

And now we can integrate both sides like this:

\int \frac{dy}{4y-3} = \int dt

We can use the u substituton for the left part u = 4y-3 du = 4 dy

\frac{1}{4} \int\frac{du}{u} = t+ c

We can multiply both sides by 4 and we got:

\int\frac{du}{4} = 4t+ 4k = 4t+ c, where c=4k

and after integrate the left part we got:

ln |u| = 4t+c

And if we apply exponential on both sides we got:

u = e^{4t} e^c

Now we can replace u and we got:

4y-3 = Ce^{4t} where C=e^c

And then finally we have:

y(t) = \frac{Ce^{4t} +3}{4}

Part 2

For this case we have the initial condition y(0) =5 and if we use it we got:

5 = \frac{C e^0 +3}{4}

20 = C +3

C= 20-3=17

So then the model would be given by:

y(t) =\frac{17te^{4t} +3}{4}

4 0
3 years ago
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