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konstantin123 [22]
3 years ago
8

Four friends share 3 orders of chicken wings Alex eats 5/8 of an order Chase eats 7/8 of an order Ella eats 6/8 of an order.How

much of an order of chicken wings is left for the fourth friend.
Mathematics
1 answer:
NikAS [45]3 years ago
3 0

Answer:

6/8

Step-by-step explanation:

5/8+7/8+6/8=2 1/4

3 - 2 1/4= 3/4

3/4 x2 =6/8

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Convert 6 20<br> into a decimal
lakkis [162]

Answer:

6.20

wait what do u mean

6 0
3 years ago
The cost of a company picnic varies directly as the number of employees attending the picnic. If a company picnic costs $487.50
Drupady [299]

The company picnic cost $ 1300 for 80 employees

<em><u>Solution:</u></em>

Given that cost of a company picnic varies directly as the number of employees attending the picnic

Let "c" be the company picnic cost

Let "n" be the number of employees attending the picnic

Therefore,

cost \propto \text{ number of employees attending picnic }

c \propto n\\\\c = kn

Where "k" is the constant of proportionality

c = kn ---------- eqn 1

<em><u>Given that company picnic costs $487.50 for 30 employees</u></em>

Therefore substitute c = 487.50 and n = 30

487.50 = k(30)\\\\k = \frac{487.50}{30} = 16.25

<em><u>How much does a company picnic cost for 80 employees?</u></em>

Substitute n = 80 and k = 16.25 in eqn 1

c = 16.25(80) = 1300

Thus $ 1300 is the cost for 80 employees

5 0
3 years ago
8)<br> 8x + 6<br> 14x - 2<br> A) 11<br> C) 8<br> B) -9<br> D) -7
Travka [436]

Answer:

C) 8

Step-by-step explanation:

Consecutive angles are equal to 180. So 22x+4+180. Solve that to get x=8. Hope I helped and post more questions :)

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
Omar has 1/4 of a pizza. He wants to eat 2/3 of this amount. How much of the pizza will omar eat ?
Talja [164]
5/12 uuuuurrr welcome

8 0
3 years ago
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