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3 years ago

What is 4 divided by -26 ?

2 answers:

5 0

You have to divide just 26 divided by 4 and that should give you 6.5 and if you have one positive and one negative the answer should be negative so the answer is -6.5

Svetlanka [38]3 years ago

4 0

<span>-0.15384615384 repeative
or -0.153846 with a line on top</span>

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What is the value of the expression 24 + 32? (2 points)

Maurinko [17]

Answer:

Step-by-step explanation:

Round: 24 is 20 and 34 is 30

20 + 30 = 50

now unrounded, it's easier.

24 + 32 = 56

because 4 + 2 = 6, we know to add on only a few more.

8 0

2 years ago

The length of one side of a square is 3n+2. The perimeter of the square is 4(3n+2). Which expression is equivalent to the perime

deff fn [24]

Answer:

4(3n + 2)

Step-by-step explanation:

as we know perimeter of a square is 4 x side

one side is 3n + 2

4 sides = 4(3n +2)

8 0

3 years ago

Can anybody help me with my work

nadya68 [22]

Answer:

$5.75

Step-by-step explanation:

14.25 divided by 5= 2.85

2.85+2.85=5.75

7 0

2 years ago

Read 2 more answers

Factoring trinomial fractions?

forsale [732]

The first one is (z-7)(z-1)
second: (x+1/3)(x+1/3)
third: (x-1/5)(x-1/5)
fourth: (x-5)(x-2)

4 0

3 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

3 years ago