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3 years ago

Answer 10,11,12,13,14

Mathematics

1 answer:

QveST [7]3 years ago

4 0

10
13.50 per set
+.45 per day value
she got 20 sets
5 days after would be
13.50 ×20=270
value of 20 sets at 13.50 a set on day one was 270.00
.45×20=9.00
the 20 sets gain 9 dollars a day
9×5=45
9 dollars a day times 5 days is 45 dollars
45+270=315

11

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Round 913610 to the nearest ten

Marizza181 [45]

The next bigger ten is 913,620 .
The next smaller ten is 913,600.

913,610 is exactly in the middle between them.

913,610 is already written to the nearest ten.

7 0

4 years ago

Read 2 more answers

The difference of two trinomials is x2 − 10x + 2. If one of the trinomials is 3x2 − 11x − 4, then which expression could be the

12345 [234]

Answer:

$4x^2-21x-2$

Step-by-step explanation:

Given that:

Difference of two trinomials is $x^2 - 10x + 2$

One of the two trinomials is $3x^2 - 11x - 4$

To find:

The other trinomial = ?

Four options are:

$2x2 - x - 2 \\2x2 + x + 6 \\4x2 + 21x + 6\\ 4x2 - 21x - 2$

Solution:

Let the two trinomials be A and B.

Given A - B = $x^2 - 10x + 2$

B = $3x^2 - 11x - 4$

We have to find the other trinomial A.

A - B = $x^2 - 10x + 2$

A - ( $3x^2 - 11x - 4$ ) = $x^2 - 10x + 2$

$\Rightarrow$ A = $x^2 - 10x + 2$ + ( $3x^2 - 11x - 4$ )

$\Rightarrow$ A = $4x^2-21x-2$

So, the correct answer is $4x^2-21x-2$ .

3 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

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