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wolverine [178]
2 years ago
15

Write a doubles fact that can help you solve 6+5=11

Mathematics
1 answer:
erma4kov [3.2K]2 years ago
3 0
6 = 5 + 1, then (5 + 1) + 5 = 11, (5 + 5) + 1 = 11, 10 + 1 = 11
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Which relation is a function?
LiRa [457]

Answer:

{(2, 3), (–2, 3), (3, 2), (–3, –2)}

Step-by-step explanation:

Functions can't have any repeating x-coordinates.

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A metal strip is being installed around a workbench that is 8 feet long and 3 feet wide. If the stripping costs $6 per foot, fin
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I believe the answer is $144
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A triangle has side lengths of 24 cm, 24 cm, and 24 cm.
zhenek [66]

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equilateral

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
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

7 0
1 year ago
Ayuda con mates por favor
QveST [7]

Answer:

a) x = -1 and  y = 5, b) x = 2 and y = -1, c) x = -2 and y = 1, d) x = 2 and y = -9, e) x = 2 and y = -2, f) x = 2 and y = -1, g) x = 2 and y = -1, h) x = 1 and y = 2.

Step-by-step explanation:

Each system is solved as follows (Cada sistema es resuelto como sigue):

a) 2\cdot x + y = 3 and 3\cdot x - y = -8

y = 3 - 2\cdot x

3\cdot x - 3 + 2\cdot x = -8

5\cdot x = -5

x = -1

y = 5

b) x - 2\cdot y = 4 and -x+3\cdot y = -5

x = 4 + 2\cdot y

-4-2\cdot y+3\cdot y = - 5

-4+y = -5

y = -1

x = 2

c) -2\cdot x + 5\cdot y = 9 and x - y = -3

x = y-3

-2\cdot y +6 +5\cdot y = 9

3\cdot y = 3

y = 1

x = -2

d) 5\cdot x - 4\cdot y = 2 and 3\cdot x + 2\cdot y = -12

3\cdot x + 2\cdot y = -12

6\cdot x + 4\cdot y = -24

4\cdot y = -6\cdot x -24

5\cdot x +6\cdot x +24 = 2

11\cdot x = -22

x = 2

y = -9

e) x + y = 0 and 2\cdot x +3\cdot y = -2

y = -x

2\cdot x -3\cdot x = -2

-x = -2

x = 2

y = -2

f) 3\cdot x + 5\cdot y = 1 and x + y = 1

y = 1 - x

3\cdot x + 5 - 5\cdot x = 1

-2\cdot x = -4

x = 2

y = -1

g) 5\cdot x - 2\cdot y = 12 and 4\cdot x + 3\cdot y = 5

x = \frac{12+2\cdot y}{5}

x = \frac{5-3\cdot y}{4}

\frac{12+2\cdot y}{5} = \frac{5-3\cdot y}{4}

4\cdot (12+2\cdot y) = 5\cdot (5-3\cdot y)

48 + 8\cdot y = 25 - 15\cdot y

23\cdot y = -23

y = -1

x = 2

h) 7\cdot x + 8\cdot y = 23 and 3\cdot x + 2\cdot y = 7

3\cdot x + 2\cdot y = 7

12\cdot x + 8\cdot y = 28

8\cdot y = 28 - 12\cdot x

7\cdot x +28-12\cdot x = 23

-5\cdot x = -5

x = 1

y = 2

8 0
3 years ago
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