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lutik1710 [3]
3 years ago
9

Diego is thinking of two positive numbers. He says, 'If we triple the first number and double the second number, the sum is 34.'

Diego then says, 'If we take half of the first number and double the second, the sum is 14.' Answer the questions in the table to model Diego's problem. Let `x` equal the first number and `y` equal the second number.
Mathematics
1 answer:
UNO [17]3 years ago
8 0

1st number: 0

2nd number: 17

Answer

Step-by-step explanation:

3x+2y=34 x is the first number, y is the second.

1/2x+2y=34 multiply each number by two to get rid of the integer by the variable x.

x+4y=68 solve for x.

x=68-4y add this into the first equation to solve for variable y.

3(68-4y)+2y=34 solve for y.

204-10y=34

-10y= -170

y=17

now to solve for x

x= 68-4(17)

x= 0

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4 0
2 years ago
I need help with this problem
Arada [10]

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65.56°

Step-by-step explanation:

We know that if we take dot product of two vectors then it is equal to the product of magnitudes of the vectors and cosine of the angle between them

That is let p and q be any two vectors and A be the angle between them

So, p·q=|p|*|q|*cosA

⇒cosA=\frac{u.v}{|u||v|}

Given u=-8i-3j and v=-8i+8j

|u|=\sqrt{(-8)^{2}+ (-3)^{2}} =8.544

|v|=\sqrt{(-8)^{2}+ (8)^{2}} =11.3137

let A be angle before u and v

therefore, cosA=\frac{u.v}{|u||v|}=\frac{(-8)*(-8)+(-3)*(8)}{8.544*11.3137} =\frac{40}{96.664}

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5 0
3 years ago
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and con
sladkih [1.3K]

65 sequences.

Lets solve the problem,

The last term is 0.

To form the first 18 terms, we must combine the following two sequences:

0-1 and 0-1-1

Any combination of these two sequences will yield a valid case in which no two 0's and no three 1's are adjacent

So we will combine identical 2-term sequences with identical 3-term sequences to yield a total of 18 terms, we get:

2x + 3y = 18

Case 1: x=9 and y=0

Number of ways to arrange 9 identical 2-term sequences = 1

Case 2: x=6 and y=2

Number of ways to arrange 6 identical 2-term sequences and 2 identical 3-term sequences =8!6!2!=28=8!6!2!=28

Case 3: x=3 and y=4

Number of ways to arrange 3 identical 2-term sequences and 4 identical 3-term sequences =7!3!4!=35=7!3!4!=35

Case 4: x=0 and y=6

Number of ways to arrange 6 identical 3-term sequences = 1

Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 1 + 28 + 35 + 1 = 65

Hence the number of sequences are 65.

Learn more about Sequences on:

brainly.com/question/12246947

#SPJ4

8 0
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