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2 years ago

Find the nth term 14,25,36,47,58,

Mathematics

2 answers:

borishaifa [10]2 years ago

8 0

Answer:

a6=69

Step-by-step explanation:

katen-ka-za [31]2 years ago

3 0

The sequence is going up by 11 each time, but the sequence starts on 14, not 11 which means the answer is: 11n + 3

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Identify the pattern for the following sequence. Find the next three terms in the sequence. 5, 2, -1, -4, ____, ____, ____,... a

Ludmilka [50]

Answer:

A. subtract s: -7, -10, -13

Step-by-step explanation:

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2 years ago

A ferris wheel can accommodate 40 in 30 minutes. how many people could ride the ferris wheel in 3 hours? what is the total rate

oksano4ka [1.4K]

In an hour, 80 people could ride the ferris wheel. So, that means in 3 hours 240 people could ride it.

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2 years ago

Whoever can help me will get the brainliest

bezimeni [28]

Answer:

Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height. plug in the numbers in the formula and get the answer for the formula

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3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

2 years ago

What is the equation of the graph below?

suter [353]

Answer:

The equation to the graph is y = − (x − 3)2 + 2.

Hope this helps!

Remember to mark me Brainliest!

6 0

2 years ago

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