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2 years ago

I NEEED HELLPPP write this using exponents question

Mathematics

1 answer:

lbvjy [14]2 years ago

5 0

Answer:

D. $(-2)^6$

Step-by-step explanation:

There are six (-2), so it would be D. since C. would just mean $2^6$ and turn that negative.

You can check using a calculator that (-2)(-2)(-2)(-2)(-2)(-2) would equal 64 while C. equals -64. So the correct answer would be D.

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Compute the exact value of the function for the given x-value without using a calculator.

avanturin [10]

Answer:

Step-by-step explanation:

Given

f(x) = 5x , then substitute x = - 1, that is

f(- 1) = 5 × - 1 = - 5 → d

3 0

3 years ago

Read 2 more answers

Quanto fa 2×2? hsbwhannwn

goldenfox [79]

Answer:

$ok, ma \: quando \: 2+2= 1? $

6 0

2 years ago

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If 3/8 of my money is $3840.00. How much did I have at first? *

NNADVOKAT [17]

Answer:

E $10240.00

Step-by-step explanation:

The total amount of money is x.

3/8 × x = 3840

Multiply both sides by the reciprocal of 3/8 which is 8/3.

x = 3840 × 8/3

x = 10240

Answer: E $10240.00

7 0

2 years ago

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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

The four angles of a quadrilateral are in the ratio 2:3:5:8. Find the angles

sp2606 [1]

Answer:

Step-by-step explanation: let 2:3:5:8 be 2x,3x,5x,8x respectively .

*angles in a quadrilateral is 360 degree

*2x+3x+5x+8x=360

*18x=360

*x=360/18

*x=20

now substitute x in these:

2x=2x20=40

3x=3x20=60

5x=5x20=100

8x=8x20= 160

so these are the following angles: 40,60,100 and 160

8 0

3 years ago