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Dennis_Churaev [7]
3 years ago
15

All numbers that end in 5 or 0 are divisible by 5 and 10 True or False?! PLZ HELP

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
5 0

All numbers that end in 5 are not divisible by 10. So this is false.

lawyer [7]3 years ago
4 0
False because 5 isnt divisible by 10 and 0 isnt divisible by anything but if you mean positive numbers with multiple digits than its true
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Hatshy [7]
If you look at the rectangle properly, you can see that the diagonal creates two triangles, and if we only focus on one triangle, we can see that the bad of the triangle is 7 and the perpendicular height is 4

Using Pythagorean theorem we can figure out the hypotenuse, or the diagonal
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3 0
3 years ago
Given the replacement set {0, 1, 2,3 4,}, solve 6x - 3=3
IrinaVladis [17]

Answer:

1

Step-by-step explanation:

6x - 3 =3

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7 0
3 years ago
A number squared times four is equal to 16. Find the missing number.
erica [24]
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6 0
3 years ago
The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

8 0
3 years ago
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garik1379 [7]

Answer:

8

Step-by-step explanation:

9-5÷(8-3)×2+6

4÷4×2+6

1×8

=8

6 0
3 years ago
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