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2 years ago

Please it’s due today I need help ASAP

Mathematics

1 answer:

yan [13]2 years ago

6 0

I think it’s B because the total he could spend is 52.50, so it could be less than that or equal to that very amount. The passage directly says he spent 31.75 on a coat. The only thing it didn’t give you a clear number on was the shirts, but it gives you the price so you put the price and the variable “s” together because you don’t know have many shirt he got.

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Boden's account has a principal of $700 and a simple

pashok25 [27]

Answer:

823.2

Step-by-step explanation:

Interest = 700 × 4.4/100 × 4 = 123.2

Balance = Principle + interest

700 + 123.2 = 823.2

Answer - 823.2

Hope this helps!!!

3 0

2 years ago

find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on

enyata [817]

Let $f(x)=\tan^{-1}x$ . Then $f'(x)=\frac1{1+x^2}$ . Note that $f(0)=0$ .

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n$

This means that for $|-x^2|=|x|^2$ , or $-1$ , we have

$\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Integrate the series to get

$f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

$\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

6 0

2 years ago

How to find slope and y intercept

Sergio [31]

Slope is rise over run

7 0

3 years ago

Read 2 more answers

If f(x)=x^3-x+2, then (f^-1)'(2)

yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

$f\left(f^{-1}(x)\right) = x$

$\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x$

which is a cubic polynomial in $f^{-1}(x)$ with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

$\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}$

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0 or x = 1 or x = -1

Then $\left(f^{-1}\right)'(2)$ can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0

1 year ago

Solve 4+2h < -3 please help me and please show work

anastassius [24]

4 + 2h < -3
2h < -7
h < -3.5

x ∈ (-∞, -3.5)

7 0

2 years ago