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gulaghasi [49]
2 years ago
10

A number is multiplied by 2,and tgen the product is added to 9. Tge result is 49.Whatvus tye number

Mathematics
1 answer:
Alex17521 [72]2 years ago
6 0

Answer:

based on your question

is this the number 20?

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What is the value of the expression given below? (8-3i)-(8-3i)(8+8i)
enyata [817]

Answer:

-76-43i

Step-by-step explanation:

First expand the multiplied terms

8-3i-(64+64i-24i+24)

8-3i-(64+40i+24)

Simplify

8-3i-64-40i-24

-76-43i

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2 years ago
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A boat is 75 miles west and 60 miles north from the nearest port. What bearing should be taken to sail directly to port? S 39° E
Nimfa-mama [501]
I dont know this is really hard

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3 years ago
True or false: Removing the parenthesis from will change the result.
egoroff_w [7]

Answer:

true

Step-by-step explanation:

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2 years ago
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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
Solve the quadratic equation by completing the square.x^2-14x+46=0First choose the appropriate form and fill in the blanks with
Kaylis [27]

x² - 14x + 46 = 0

subtract 46 from both-side

x² - 14x = -46

Add the square of the half of the co-efficient of x

x² - 14x + (-7)² = -46 + 7²

(x-7)² = -46 + 49

(x-7)² =3

Take the square root of both-side

x-7 = ±√3

x-7= ±1.732

Add 7 to both-side of the equation.

x= 7 ± 1.732

Eithe x= 7 + 1.732 or x= 7 - 1.732

x=8.732 or x=5.268

Therefore x = 8.732 , 5.268

7 0
1 year ago
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