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3 years ago

If F(x) = 2x-3/5 which of the following is the inverse of F(x)?

Mathematics

1 answer:

fredd [130]3 years ago

3 0

The inverse would be f^-1 (x) = 1/2x + 3/10

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How to I find a? It’s round to the nearest tenth.

Irina18 [472]

Answer:

a ≈ 21.8

Step-by-step explanation:

We require the third angle in the triangle

Subtract the sum of the 2 angles from 180

third angle = 180° - (75 + 31.8)° = 180° - 106.8° = 73.2°

Using the Sine rule, that is

$\frac{a}{sin73.2}$ = $\frac{22}{sin75}$ ( cross- multiply )

a × sin75° = 22 × sin73.2° ( divide both sides by sin75° )

a = $\frac{22sin73.2}{sin75}$ ≈ 21.8 ( to the nearest tenth )

7 0

4 years ago

This is the other question!

joja [24]

I’m not 100% true but i think it’s false, true, true :)

6 0

2 years ago

Read 2 more answers

Which expression is equivalent to

LiRa [457]

Answer: B 1/m

Step-by-step explanation:

7 0

3 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$

$\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4$

$\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4$

$\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4$

Substituting the values in the series expansion gives:

$e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...$

Factoring out e⁴:

$e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]$

<u>Taylor Series summation notation</u>:

$\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n$

Therefore:

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

7 0

2 years ago

What does x and y =??????

Kazeer [188]

Answer:

x=30. y=60

Step-by-step explanation:

DG is a straight segment so it sums to 180 degrees. ECG is a right angle so 90 degrees. The rest is described as 2x and x. So 3x=90 degrees. x=30. Eh is also a straight segment. since we know x, we know that ECD is 60 degrees. So y+y=120 degrees. 2y=120. y=60.

6 0

3 years ago