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DENIUS [597]
3 years ago
10

Can someone help me quick please

Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0

Answer:

30/7 --> 4 2/7

Step-by-step explanation:

2 1/7 x 3 2/3

Katena32 [7]3 years ago
6 0

Answer: 11 619/1000

Step-by-step explanation:

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Which description best describes the solution to the following system of equations?
sertanlavr [38]
Line y = –x + 4 intersects the line y = 3x + 3., this is the right answer
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Explain and answer please throughlly explain. I got 7 with a remainder of 3 halp. I think its .73 idk.
kipiarov [429]

Answer:

H) 0.75

Step-by-step explanation:

4.5 divided by 6 is 0.75

We don't use remainders in algebra for the most part.  We have to do regular division.  I'll try to show it here:

   

   ____._7___5_____________

6 ) 4    .    5     0

    4    .     2

 ____________

    0   .     3    0

    0  .      3    0

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3 years ago
If f(x)=4x+9+2, which inequality can be used to find the domain of f(x
Ksju [112]
First off, the function above is incomplete, however, by the looks of it, likely a quadratic, anyway, the values "x" can safely take on would be all real numbers, thus  -∞ < x < +∞
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}

3 0
3 years ago
Bess has $2.80 in quarters and dimes. The number of dimes is 7 less than the number of quarters. Find the number of quarters and
Nezavi [6.7K]

Answer:

3 dimes and 10 quarters

Step-by-step explanation:

If you add 7 dimes, then the number of dimes and quarters is the same, and the total increases to $3.50. A dime and a quarter have a value together of $0.35. Since there are equal numbers, the value $3.50 must be made up of some quantity of groups with a value of $0.35. That number of groups must be ...

... $3.50/$0.35 = 10

Thus, there are 10 quarters and 10-7 = 3 dimes.

_____

<em>Using an equation</em>

Let q represent the number of quarters. Then the amount Bess has is ...

... 0.10(q -7) +0.25q = 2.80

... 0.35q - 0.70 = 2.80

... 0.35q = 3.50 . . . . . add 0.70 . . . . . . . this should look familiar

... q = 3.50/0.35 = 10

4 0
3 years ago
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