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3 years ago

Pls help asap i will give branerlist plus 10 ponts

Mathematics

2 answers:

Luden [163]3 years ago

8 0

4 and 5 are the only constant terms

sasho [114]3 years ago

6 0

Answer:

This is what I got 13xyz+2z+9

Step-by-step explanation:

Hope that helps

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This is a tough question.

tiny-mole [99]

Y is greater than or equal to 0

8 0

2 years ago

Read 2 more answers

9. Solve the system of equations using substitution. y = 2x - 10 y = 4x - 8

krok68 [10]

Since they’re both in standard form and they both say that y is equal to something, you just have to set them up with one another
2x-10=4x-8
Subtract 2x
-10=2x-8
Add 8
-2=2x
Divide by 2
-1=x
x=-1
Check it by inserting it
2(-1)-10=4(-1)-8
-2-10=-4-8
-12=-12
So x=-1 is the answer

3 0

3 years ago

Read 2 more answers

Given the sequence, 26, 13, 6.5, ..... find a) the 10th term and b) the sum of the first 18 terms

Hunter-Best [27]

Answer:

a) 10th term is 0.051

b) The sum of first 18 terms of given sequence is 51.48

Step-by-step explanation:

We are given the sequence 26, 13, 6.5, ..... we need to find

a) 10th term

b) Sum of first 18 terms

Before solving we need to determine if the sequence is arithmetic or geometric

The sequence is arithmetic if common difference d is same.

The sequence is geometric if common ratio r is same.

Finding common difference d: 13-26 = -13, 6.5-13= -6.5

As common difference is not same so, the sequence is not arithmetic.

Finding common ratio r : 13/26 =0.5, 6.5/13= 0.5

As common ratio is same so, the sequence is geometric.

a) Finding common difference: 13-26 = -13, 6.5-13= -6.5

As common difference is not same so, the sequence is not arithmetic.

a) 10th term

The formula to find 10th term is: $a_n=a_1r^{n-1}$

We have a₁=26 and r = 0.5 n=10

$a_n=a_1r^{n-1}\\a_{10}=26(0.5)^{10-1}\\a_{10}=26(0.5)^{9}\\a_{10}=0.051$

So, 10th term is 0.051

b) Sum of first 18 terms

The formula to find sum of geometric series is: $S_n=\frac{a(1-r^n)}{1-r}$

where a= 1st term, r = common ratio and n= number of terms

In the given sequence we have

a=26, r=0.5 and n=18

Finding sum of first 18 terms

$S_n=\frac{a(1-r^n)}{1-r}\\S_{18}=\frac{26(1-(0.5)^{18}}{1-0.5}\\S_{18}=\frac{26(0.99)}{0.5}\\S_{18}=\frac{25.74}{0.5}\\S_{18}=51.48$

So, sum of first 18 terms of given sequence is 51.48

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:

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}}$