All categories

3 years ago

I need help plz! If anyone knows, I would truly appreciate it! Thank uuu!!!!^^

Mathematics

2 answers:

swat323 years ago

8 0

Answer:

pic1 a pic2 d

Step-by-step explanation:

i did this

IrinaVladis [17]3 years ago

4 0

1 : D 2 : A

i hade this paper earlier

You might be interested in

Please help me this is due tomorrow and I just need this. ALSO WHEN YOU ANSWER SHOW ME YOUR STEPS FOR BRAINLEST THANKS!

Gnesinka [82]

Answer:

x=3

Step-by-step explanation:

1/4(12x + 4) - 14= -1/2(8x - 16)

1. Simplify each side my multiplying

3x - 13 = -4x + 8

2. Add 13 on each side

3x = -4x + 21

3. Add 4x on each side

7x = 21

x=3

6 0

3 years ago

HELP ASAP

GarryVolchara [31]

Answer:

Todd paid $0.08 less per gallon of gasoline than Hope.

4 0

3 years ago

Read 2 more answers

Write an algebraic rule to describe the translation C(5, –4)- ’(–2, 1).

Orlov [11]

A negative minus a negative is always a positive ;)

5 0

2 years ago

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

2 years ago

Which of the following numbers are greater than or equal to 4/7

Alexandra [31]

B and C mark brainly!

8 0

2 years ago