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3 years ago

Can someone help me with this pls no bull sh answers

Mathematics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

(x-1)(-3x-7)

Step-by-step explanation:

To find this answer I used this method:

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Can someone help me with this ASAP please I’m being timed !

sergij07 [2.7K]

i really cant see what it says sorry

7 0

3 years ago

Two passenger train, A and B, 450 km apart, star to move toward each other at the same time and meet after 2hours.

likoan [24]

Let $v$ be the speed of train A, and let's set the origin in the initial position of train A. The equations of motion are

$\begin{cases}s_A(t) = vt\\s_B(t) = -\dfrac{8}{7}vt+450\end{cases}$

where $s_A,\ s_B$ are the positions of trains A and B respectively, and t is the time in hours.

The two trains meet if and only if $s_A=s_B$ , and we know that this happens after two hours, i.e. at $t=2$

$\begin{cases}s_A(2) = 2v\\s_B(2) = -\dfrac{16}{7}v+450\end{cases}\implies 2v = -\dfrac{16}{7}v+450$

Solving this equation for v we have

$2v = -\dfrac{16}{7}v+450 \iff \dfrac{30}{7}v=450 \iff v=\dfrac{450\cdot 7}{30} = 105$

So, train A is travelling at 105 km/h. This implies that train B travels at

$105\cdot \dfrac{8}{7} = 15\cdot 8=120 \text{ km/h}$

5 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

Jessica Barth 15 pieces of ribbon each piece of ribbon was 2.25 yards long what is the total amount of ribbon Jessica bu Jessica

Flauer [41]

The answer is 3.375

6 0

3 years ago

Find the balance on a deposit of $455 that earns 4% interest compounded annually for 2 years. $36.40 $491.40 $492.13 $819

STatiana [176]

$A=P(1+ \frac{r}{n})^{tn}$
A=future amount
P=present amount
r=rate in decimal
n=number of times per year compounded
t=time in years

given
P=455
r=4%=0.04
n=1
t=2

$A=455(1+ \frac{0.04}{1})^{(2)(1)}$
$A=455(1+ 0.04)^{2}$
$A=455(1.04)^{2}$
A=492.128
round
$492.13
3rd option

5 0

4 years ago

Read 2 more answers