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FromTheMoon [43]
3 years ago
5

Which of the following is not a real number?

Mathematics
1 answer:
Aleks [24]3 years ago
3 0

Answer:

D:  √(-4)

Step-by-step explanation:

D:  √(-4) is not a real number; it's an imaginary one.

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1 3/4 − 4/5 = 35 twentieths − what fraction???<br><br>Plss help will give brainliest :)
7nadin3 [17]

let's firstly convert the mixed fractions to improper fractions and then to do away with the denominators, let's multiply both sides by the LCD of all denominators.

\stackrel{mixed}{1\frac{3}{4}}\implies \cfrac{1\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{7}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{4}-\cfrac{4}{5}=\cfrac{35}{20}-\boxed{?}\implies \stackrel{\textit{multipling both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{7}{4}-\cfrac{4}{5} \right)=20\left( \cfrac{35}{20}-\boxed{?} \right)} \\\\\\ 35-16=35-20\boxed{?}\implies 19=35-20\boxed{?}\implies -16=-20\boxed{?} \\\\\\ \cfrac{-16}{-20}=\boxed{?}\implies \cfrac{4}{5}=\boxed{?}

8 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!
Montano1993 [528]

Answer:

Yes. In general, the set of rational numbers is closed under addition, subtraction, and multiplication; and the set of rational numbers without zero is closed under division.

Step-by-step explanation:

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3 0
3 years ago
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Please help ASAP need answer fast
kap26 [50]
A. 2
b. 0
c. -2
d. 0
f(x) just means that the number in the parentheses is the x-value and if it asks for f(x) it wants the y-value at that x point
5 0
3 years ago
What is the center of the circle shown below?
ExtremeBDS [4]
The center is M.
Your answer is B.
5 0
3 years ago
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Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e
Bingel [31]
Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0

and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
6 0
3 years ago
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