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1 year ago

Can someone please help

Mathematics

1 answer:

Sloan [31]1 year ago

7 0

Answer:

Step-by-step explanation:

an = a1 + (n-1)d

a_n = the nᵗʰ term in the sequence = 10

a_1 = the first term in the sequence = 5

d = the common difference between terms = 4

a10 = 5 + (10-1)4

a10 = 5 + (9)4

a10 = 5 + 36

a10 = 41

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What is this number in standard form? (3×100)+(1×110)+(4×11,000)

zimovet [89]

Answer:

44410

Step-by-step explanation:

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2 years ago

Find the average rate of change of f(x)=2x-1 from x=-1 to x=2

MrMuchimi

The average rate of change from x = -1 to x = 2 is 2

Solution:

Given function is:

f(x) = 2x - 1

We have to find the average rate of change from x = -1 to x = 2

The average rate of change is given as:

$\text {Average rate of change}=\frac{f(b)-f(a)}{b-a}$

The average rate of change from x = -1 to x = 2 is given by formula:

$\text {Average rate of change}=\frac{f(2)-f(-1)}{2-(-1)}\\\\\text {Average rate of change}=\frac{f(2)-f(-1)}{3}$

Find f(2) and f( - 1)

Substitute x = 2 in given function

f(2) = 2(2) - 1 = 4 - 1 = 3

Substitute x = -1 in given function

f( - 1) = 2(-1) - 1 = -2 - 1 = -3

Substitute the values in above formula,

$\text {Average rate of change}=\frac{3-(-3)}{3}=\frac{6}{3}=2$

Thus average rate of change from x = -1 to x = 2 is 2

3 0

2 years ago

City L has a temperature of −3 °F. City M has a temperature of −7 °F. Use the number line shown to answer the questions: Number

Alik [6]

-3 > -7
u have to remember, with negatives, the larger number is smaller in value.

when u plot these numbers on the number line, the number that is farthest to the left is less then...the number that is farthest to the right is greater then.

-3 is warmer because it is closer to zero then -7.

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2 years ago

A while back, either James borrowed $12 from his friend Rita or she borrowed $12 from him, but he can’t quite remember which. Ei

Ket [755]

He will have 32.00 left from paying back rita.

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3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using integration by parts:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

1 year ago