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3 years ago

Which are zeros of the function represented by this graph?

Mathematics

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

x = - 2 and x = 4

Step-by-step explanation:

The zeros of the function are the values of x where the graph crosses the x- axis ( also called the x- intercepts )

The graph crosses the x- axis at

x = - 2 and x = 4 ← zeros

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There are 346 children going for an excursion.

NeTakaya

Answer:

26 children

Step-by-step explanation:

346-8*40

346-320

5 0

2 years ago

How do you simplify odd fractions?

gtnhenbr [62]

Answer:

Step-by-step explanation:

How do you simplify fractions with odd numbers? Find the greatest common factor of both the numerator and denominator. Divide the numerator by the GCF. The result will be the simplified numerator.

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

1 year ago

3.11 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the nu

djverab [1.8K]

Answer:

Probability distribution for x:

$P(x=0)=0.3644\\\\P(x=1)=0.4373\\\\P(x=2)=0.1749\\\\P(x=3)=0.0233\\\\$

Step-by-step explanation:

We can model the number of defective sets in the group of TV sets (variable x) as a binomial variable, with sample size=3 and probability of success p=2/7≈0.2857.

The probability of k defective sets in the group is:

$P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}$

So, we have this probabilty distribution for x:

$P(x=0) = \dbinom{3}{0} p^{0}q^{3}=1*1*0.3644=0.3644\\\\\\P(x=1) = \dbinom{3}{1} p^{1}q^{2}=3*0.2857*0.5102=0.4373\\\\\\P(x=2) = \dbinom{3}{2} p^{2}q^{1}=3*0.0816*0.7143=0.1749\\\\\\P(x=3) = \dbinom{3}{3} p^{3}q^{0}=1*0.0233*1=0.0233\\\\\\$