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NISA [10]
3 years ago
11

How to solve three number questions?

Mathematics
1 answer:
umka2103 [35]3 years ago
8 0
What numbers are u talking about
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Y = x - 1
romanna [79]

Answer: The correct answer is the third option, (-3,-4).

Step-by-step explanation: You plug in 2x + 2 for y, given that y = 2x + 2.

2x+2 = x - 1

x = -3

Now, you plug in -3 for x in an equation.

y = -3 - 1

y = -4.

(-3, -4)

4 0
2 years ago
Read 2 more answers
If the binomial 2x+8 was multiplied by 5 the result would be equivalent to
ivolga24 [154]

Answer:

(1) 10x + 40

Step-by-step explanation:

5(2x+8)

= 5×2x + 5×8

= <u>10x + 40 (Ans)</u>

5 0
3 years ago
A 11 ft board is to be cut into three​ pieces, two​ equal-length ones and the third 3 in. shorter than each of the other two. if
KonstantinChe [14]
11 feet = 132 inches
The boards are x, x and (x-3) inches long.
x + x + x -3 = 132
3x = 135
x = 45
The boards are 45 inches, 45 inches and 42 inches long.


6 0
4 years ago
Which shape has 2 parallel sides /i ready question
sesenic [268]

Answer:

A Parallelogram

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Let a "binary code" be the set of all binary words, each consisting of 7 bits (i.e., 0 or 1 digits). For example, 0110110 is a c
Dmitry_Shevchenko [17]

Answer:

a) 128 codewords

b) 35 codewords

c) 29 codewords

Step-by-step explanation:

a) Each 7 bits consist of 0 or 1 digits. Therefore the first bit is two choices (0 or 1), the second bit is also two choices (0 or 1), continues this way till the last bit.

So total number of different code words in 7 bits is 2×2×2×2×2×2×2 = 2⁷ = 128

There are 128 different codewords.

b) A code word contains exactly four 1's this means that  it has four 1's and three 0's . Therefore, in 7 bits, we have four of the same kind and three of the same kind. Hence, total number of code words containing exactly four 1's =7!/(4!*3!) = 35 codewords

c) number of code words containing at most two 1's  = codewords containing zero 1's + words containing one 1's + words containing two 1's

Now codewords containing zero 1's = 0000000 so 1 word

Codewords containing one 1's = 1000000,0100000,0010000,0001000,0000100,0000010,0000001. That's seven words

Codewords containing two 1's means word containing two 1's and five 0's. So out of seven, two are of one kind and five are of another kind

Therefore, the total number of such words=7!/(2!*5!)=21

Hence, codewords having at most two 1's = 21+7+1 =29 codewords

8 0
3 years ago
Read 2 more answers
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