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3 years ago

Heres the lineeeeeeeeeeeeeeeeeeeeee

Mathematics

1 answer:

ra1l [238]3 years ago

6 0

Answer:

Z, K and R

and

H ,B and C

........

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Create an exponential function that starts at 650 and increases by 12% us the form f(x) = p (1=r)^x

ololo11 [35]

F(x)= 650(1+.12)^x

f(x) = 650(1.12)^x

4 0

3 years ago

What is your favorite marvel character?

aliina [53]

Answer:

nico

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

F(x) = 2x2 + 2x2-,

astra-53 [7]

Answer: A

Step-by-step explanation:

A function is the comparing of two numbers all this is doing is multiplying two numbers.

3 0

4 years ago

Please help I've been trying so many times

Annette [7]

Answer:

A. -2, 1

Step-by-step explanation:

Observe the chart and see where the f(x) and g(x) intersect.

When x=-2, f(-2)=g=(-2) and when x=1, f(1)=g=(1).

7 0

3 years ago