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alina1380 [7]
2 years ago
6

If x = 2 and y = -3, what is the value of 2x^2 + 3xy – 2y^2?

Mathematics
1 answer:
sammy [17]2 years ago
7 0

Answer: Given as,X=2 & Y=3,

now multiplying we get,

2xxy=2×2×3=12

Ans.) 12

Step-by-step explanation:

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Use some of the digits 1,3,5,7,9 only once to write 5 numbers less than 2000
Yuri [45]

Answer:

1357, 1359, 1579, 1573, 1375, 1379, 1957, 1953

Step-by-step explanation:

Since this number needs to be less than 2000, the first number has to be 1. You can randomly pick and choose any of the numbers after that.

3 0
3 years ago
Write the ratio of the area of a circle with radius r to the circumference of the same circle
Goshia [24]

Answer:

\frac{r}{2\pi r} =\frac{1}{2\pi }

Step-by-step explanation:

A ratio is a comparison of two quantities and can be written in several forms including fractions. It is most commonly written in fraction form or a:b.

To write a ratio, we count the number of each quantity we are comparing or use the variable for that quantity. We write radius:circumference. Recall, the circumference of a circle can be found using \pi d or 2\pi r.

We write r: \pi d  or r:2\pi r.

We can also write in fraction form:

\frac{r}{\pi d} or \frac{r}{2\pi r} =\frac{1}{2\pi }


6 0
3 years ago
X - 6 = 5x – 2<br> help
KiRa [710]

Answer:

-6+2=5x-x

-4=4x

-4/4=X

0=x

8 0
3 years ago
A circle has a circumference of 7.85 yards what is the diameter of the well
Vsevolod [243]
2.5 yds
C=2πr
d=2r
Solving ford
d=C
π=7.85
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4 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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