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Eddi Din [679]
2 years ago
7

Factor the expression below 36a^2-25b^2

Mathematics
2 answers:
Delicious77 [7]2 years ago
6 0

Answer:

6a+5b

Step-by-step explanation:

just olya [345]2 years ago
3 0

Answer:

(6a + 5b)(6a - 5b)

Step-by-step explanation:

Rewrite the expression in the form of a² - b²:

(6a)² - (5b)²

Use the difference of squares (a² - b² = (a + b)(a - b)):

(6a + 5b)(6a - 5b)

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May someone please help me on which to graph?
olga_2 [115]

Answer:

Option (4)

Step-by-step explanation:

Proportional relationship means,

y ∝ x

y = kx

k=\frac{y}{x}

Here, k = proportionality constant

Therefore, if the graph of a line passes through the origin (0, 0) table will represent the proportional relationship.

From table 1,

For a point (1, 2)

k=\frac{2}{1}=2

For another point (3, 2)

k=\frac{3}{2}=1.5

In both the cases 'k' is not same of constant.

Therefore, table (1) is not proportional.

For table (2),

Line passes through (2, 0).

That means there is a x-intercept → (2, 0)

Therefore, table doesn't represent a proportional relationship.

For table (3),

Line passes through a point (0, 1)

It means given line has a y-intercept → y = 1

Therefore, table doesn't represent a proportional relationship.

For table (4),

Line of this table passes through two points (1, 3) and (2, 6)

k=\frac{3}{1}=3

k=\frac{6}{2}=3

Therefore, proportionality constant for the given table is 3.

Now we can graph table (4).

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2 years ago
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31.4 bc 3.14 x 2 x 5
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7 0
3 years ago
WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"
Feliz [49]

Recall Euler's theorem: if \gcd(a,n) = 1, then

a^{\phi(n)} \equiv 1 \pmod n

where \phi is Euler's totient function.

We have \gcd(9,32) = 1 - in fact, \gcd(9,32^k)=1 for any k\in\Bbb N since 9=3^2 and 32=2^5 share no common divisors - as well as \phi(9) = 6.

Now,

37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} =  32\cdot\left(32^{(\cdots)}\right)^6

where the c_i are positive integer coefficients from the binomial expansion. By Euler's theorem,

\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9

so that

32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9

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1 year ago
Brainlist! in what ways could the shape pattern below be described?​
creativ13 [48]
A . the 13th spot will have a star
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