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tamaranim1 [39]
3 years ago
9

Help plzzzzzz.........

Mathematics
1 answer:
liubo4ka [24]3 years ago
3 0

Answer:

quotient= 12

divisor= 3

divident= 36

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Which sets are not closed under subtraction
anastassius [24]

If you're referring to the different sets of real numbers, it's the ones that you could try to do subtraction and not get an answer that still in that set.

For example, natural numbers (aka 1, 2, 3, 4, ...) are not, because 7 - 11 = -4 and -4 is not a natural number.

Also, whole numbers (aka 0, 1, 2, 3, 4, ...) has the same issue.

Basically any set of real numbers that doesn't include negative numbers will have this issue.

6 0
3 years ago
Circle the expressions that represent 10% of the original cost
Anni [7]

Answer:

What are the choices?

Step-by-step explanation:


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4 years ago
Find the equation and solve it.
Mice21 [21]
To solve this, you divide 116 by four, this gets you 29. The loser got 29 votes. Then you subtract 116-29 which gives you 87. That is how many votes the winner received. Hope this helps
8 0
3 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
2 years ago
A spinner has ten equal sectors labeled from 1 to 10. The spinner is spun once.
trapecia [35]

Answer:

With probability you can multiply your chances in this case.

Getting an odd number: 1/2

Getting a number greater than 4: 1/6

Therefore it is a 1/12 chance to be both odd and greater than 4

8 0
3 years ago
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