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3 years ago

Please help! Correct answers only please!

Mathematics

1 answer:

Marrrta [24]3 years ago

3 0

Answer:

D. is your answer

Step-by-step explanation:

$V=whl=7*6.5*7=318.5unit^3$

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What is the value of x in the eqation 5[x-6] = 2 [x+3]

topjm [15]

Answer:

x = 12

Step-by-step explanation:

5x - 30 = 2x + 6

3x = 36

x = 12

8 0

3 years ago

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A lamp was marked up 100% from an original purchase cost of $30. Yesterday, Suzie sold the lamp, earning a 20% commission on the

xz_007 [3.2K]

100% markup means the price is double what it was bought for, $30. 30x2=60 dollar sale price. A 20% commision would be $12. 60 x 0.2=12

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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

Factorization for the monomial<br> <img src="https://tex.z-dn.net/?f=48x%5E%7B10%7D" id="TexFormula1" title="48x^{10}" alt="48x^

Schach [20]

Answer:

Some of the possible factorizations of the monomial given are:

$(16x^5)(3x^5)$

$(48x)(x^9)\\\\(4x^5)(12x^5)\\\\(16x^2)(3x^8)\\\\(2x^2)(4x^7)(6x)$

Step-by-step explanation:

To factorize the monomia you need to express it as a product of two or more monomials. Therefore, you must apply the proccedure shown below:

- Descompose into prime numbers:

$48x^{10}=2*2*2*2*3*x*x*x*x*x*x*x*x*x*x$

- Then, keeping on mind that, according to the Product of powers property, when you have two powers with equal base you must add the exponents, you can make several factorizations. Below are shown some of the possible factorizations of the monomial given:

$(2^4x^5)(3x^5)$

$(48x)(x^9)\\\\(4x^5)(12x^5)\\\\(16x^2)(3x^8)\\\\(2x^2)(4x^7)(6x)$

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3 years ago

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Robert found jeans on sale for 20%off the jeans were originally 65$ how much did he pay for the jeans

Mademuasel [1]

20% of 65 is $52

Hope this helps

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3 years ago

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