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Monica [59]
3 years ago
6

What is a squared 9 in algebraic form?

Mathematics
1 answer:
GrogVix [38]3 years ago
7 0
ANSWER: 9 squared in algebraic form is 9^2
9^2 equals 81
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It's in the attachment <br> URGENT
maxonik [38]

Step-by-step explanation:

\frac{2a}{a + b - c}  =  \frac{2b}{b + c - a}  =  \frac{2c}{a + c - b}  = k \\  \\ by \: theorem \: on \: equal \: ratios \\   \\ \frac{2a + 2b + 2c}{a + b - c + b + c - a + a + c - b}  = k \\  \\  \frac{2(a + b + c)}{a + b + c}  = k \\ \\   \therefore \: k = 2

5 0
3 years ago
In a woodworking class, you need to make five triangles out of plywood. Each triangle has a height of 6 inches and a base of 14
Bogdan [553]

Answer:5 * (1/2 * 6 * 14)= x

Step-by-step explanation: :D

3 0
3 years ago
How do i find the roots of this equation: f(x) = 2x5 - 9x4 + 12x3 - 12x2 + 10x - 3 = 0
Fantom [35]
The equation is <span>f(x) = 2x5 - 9x4 + 12x3 - 12x2 + 10x - 3 = 0
remark:
the sum of coefficients is = 2-9+12-12+10-3=-7+7=0, so a=1 is a zero of f
f can be written as </span>f(x) = (x-1)Q(x), and f(x) / (x-1)= Q(x)
after euclid's division
Q(x) = 2x4 - 7x3 + 5x2 - 7x +3
so for finding the other roots, making Q(x) for product of factors is a must
4 0
3 years ago
<img src="https://tex.z-dn.net/?f=42%20%5Ctimes%2042%20%5Ctimes%2029%20%3D%20" id="TexFormula1" title="42 \times 42 \times 29 =
Blababa [14]
The correct answer is 51156



Hope this helps :)
6 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
2 years ago
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