Name a number that is an integer, but not whole number.

Mathematics

1 answer:

tekilochka [14]2 years ago

4 0

Answer:

-1

Step-by-step explanation:

All negative integers are not whole numbers

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Donte simplified the expression below.

sergij07 [2.7K]

The answer is: [A]: He did not apply the distributive property correctly for 4(1 + 3i) .
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Explanation:
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Note the distributive property of multiplication:
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a*(b+c) = ab + ac.
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As such: 4*(1 + 3i) = (4*1) + (4*3i) = 4 + 12i ;
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Instead, Donte somehow incorrectly calculated:
_____________________________________
4*(1 + 3i) = (4*1) + 3i = 4 + 31; (and did the rest of the problem correctly);

Note: - (8 - 5i) = -8 + 5i (done correctly;
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So if Donte did not apply the distributive property correctly for 4*(1+3i)—and incorrect got 4 + 3i (as mentioned above); but did the rest of the problem correctly, he would have got:
_____________________________
4+ 3i - 8 + 5i = -4 + 8i (the incorrect answer as stated in our original problem.
__________________
This corresponds to: "Answer choice: [A]: He did not apply the distributive property correctly for 4(1 + 3i)."
___________________________

6 0

3 years ago

Read 2 more answers

Solve for f d=16ef^2

vitfil [10]

Here are a bunch of CORRECT answers. Your answer is in the first pic. I got number 3 wrong, but it still showed the correct answer.

6 0

3 years ago

Write a system of two equations in two variables to solve the problem. when fully extended, a ladder is 28 feet in length. if th

Alisiya [41]

Let the base of the ladder be x ft and the extension be y ft.

"when fully extended, a ladder is 28 feet in length."

means: x+y=28

"the extension is 4 feet shorter than the base"

means: y=x-4, that is x-y = 4, (taking y to the side of x, and -4 to the other side of x and y)

thus, we have the system of equations:

1) x+y= 28

2) x-y=4

adding the 2 equations, we have:

(x+y)+(x-y)=28+4

2x=32

x=32/2=16

x+y=28, so 16+y=28, thus y=12

Answer: Base: 16, Extension 12

5 0

2 years ago

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$