Two examples of hundreds place

Solving system by graphing y= -5/3x+3 y=1/3x-3

choli [55]

the point of intersection on the graph are (3,-2)

4 0

2 years ago

Select all of the numbers that are prime 17 18 19 20 21 22 23

Sveta_85 [38]

17, 19, 23.

Remember, prime numbers are numbers that can only be divided by 1 and itself.

8 0

2 years ago

Read 2 more answers

Work out m and c for the line: 3 y + 6 x = − 3

Usimov [2.4K]

Answer:

m is -2 and c is -1

Step-by-step explanation:

• Let's first phrase out the general equation of a line

${ \rm{y = mx + c}}$

m is the slope
c is the y-intercept

[ remember that a general line equation must be in slope - intercept form as shown above ]

• from our question, we are given the equation;

${ \rm{3y + 6x = - 3}}$

• let's make y the subject in order to make the equation in slope - intercept format.

→ remember to apply "subject making knowledge"

${ \rm{3y = - 3 - 6x}} \\ \\ { \rm{3y = - 6x - 3}} \\ \\ { \rm{ \frac{3y}{3} = \frac{ - 6x}{3} - \frac{3}{3} }} \\ \\ { \boxed{ \rm{y = - 2x - 1}}}$

• The above boxed equation is now a general equation. Let's extract out slope, m and y-intercept, c

${ \rm{m \: \dashrightarrow \: - 2}} \\ \\ { \rm{c \: \dashrightarrow \: - 1}}$

8 0

2 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:

Taylor series 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:

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

Taylor Series summation notation:

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

Therefore:

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

7 0

1 year ago

Why does stock have more potential for higher returns than bonds?

morpeh [17]

Answer:

Stocks have historically delivered higher returns than bonds because there is a greater risk that, if the company fails, all of the stockholders' investment will be lost.

Step-by-step explanation:

Hope this will help

7 0

3 years ago