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lisabon 2012 [21]
2 years ago
6

Find the slope of a line parallel to the line whose equation is 3Y- 5x =15

Mathematics
1 answer:
Vanyuwa [196]2 years ago
4 0

Answer:

The slope is 5/3

Step-by-step explanation:

3y-5x=15 turned into a slope intercept form is y=5/3x+5.

5/3 is the slope part of this equation.

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The measured length is 250 ft. The actual length is 246.9 ft. Find the percent error.
Marta_Voda [28]
Percent error = 1.2556%

Steps:
Percent Error =
Vobserved - Vtrue
Vtrue
=
250 - 246.9
246.9
=
3.1
246.9
= 1.2555690562981%

7 0
3 years ago
Solve for f(2)<br>f(x) =-4x+3<br>f(2)=-4(2)+3<br>f(2)=?​
Mila [183]
The answer would be -11
5 0
3 years ago
parents volunteer to make the cones they are given a recipe the recipe calls for 61/2 cups of flour duyi has already added 23/4
frozen [14]
6 1/2 - 2 3/4 = 3 3/4
6.5 - 2.75= 3.75
6 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
ANSWER QUICKLY PLEASE<br> Given the triangles are similar, find BD.
Alla [95]

Answer:

The answer is 10

Step-by-step explanation:

Trust me

5 0
3 years ago
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