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scoray [572]
3 years ago
13

What is 10 to the power of 3 multiply by 17.55?

Mathematics
2 answers:
Alexus [3.1K]3 years ago
4 0


10³ = 1 000

17.55 x 1 000 = 17550


MrMuchimi3 years ago
4 0
(10^3)*17.55=17550

17550 is your answer
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Two fractions are given. For each one, write its decimal equivalent and determine if the decimal is terminating or non terminati
ziro4ka [17]

Answer:

9/20 yes, 4/15 not. See below

Step-by-step explanation:

Pick 9/20 and multiply numerator and denominator by 5:

9/20 = 45/100

We know that if we divide a number by 100 we need to move the coma as two places left, so:

9/20 = 45/100 = 0.45

And this is a terminal decimal as we know where it ends.

On the other hand if we pick 4/15 let try to divide it (here I will do it 'manually'):

4 |_ 15

we can divide 4 by 15, so we use 40 and begin with a comma

40 |_ 15

      0.

15 enters 2 times in 40 with a rest of 10, so:

40 |_ 15

  30  0.2

  100

100 divided by 15 is 6 and we have 10 as rest again, and again and again...

40 |_ 15

  30  0.266.....

  100

    100

      ....

So, we will have 0.266666666666666 infinitely. The decimal for 4/15 is non terminating and is 0.26666666666666666...

6 0
3 years ago
I need help plzzz it is due in 5 minutes
gladu [14]

Answer:

Just tried this on a calculator

)) i think its 7/2!

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
If you had to add two fractions with denominators of 4 and 5. What might the denominator of the sum be? Show more than one possi
mylen [45]

Answer:

your answer would be 20 because you would find what both can go into.

5 0
3 years ago
If I put 1500 into my savings account and 180 of interest at 4% simple interest how long was my money in my bank
stealth61 [152]

10,800 I think that's the answer


6 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
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