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elena-14-01-66 [18.8K]
3 years ago
15

I don’t understand where to start or what to do , help pls .

Mathematics
2 answers:
7nadin3 [17]3 years ago
5 0

Answer:

x = -19

Step-by-step explanation:

For this problem, we simply need to get rid of the fractions so we can easily solve for x.

To get rid of these fractions, we will multiply the whole equation by the LCD which is 2 * 3 = 6.

( x / 2) - ( 1 / 2) = ( 2x / 3) + ( 8 / 3 )

6 * ( x / 2) - 6 * ( 1 / 2) = 6 * (2x / 3) + 6 * ( 8 / 3 )

3x - 3 = 4x + 16

3x - 3x - 3 - 16 = 4x - 3x + 16 - 16

0x - 19 = x + 0

x = - 19

Thus, for the equation to be true, x must be equal to negative 19.

Cheers.

My name is Ann [436]3 years ago
4 0

Answer:

Answer= -19

Step-by-step explanation:

x/2 - 1/2 = 2x/3 + 8/3

x-1 / 2 = 2x+8 / 3

(x - 1) × 3 = (2x + 8) × 2

3x - 3 = 4x + 16

3x - 4x = 16 + 3

-x = 19

x = -19.

Thus, the value of x will be -19.

Hope it helps.

Mark it as Brainliest.

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Find two additional roots of P(x)=0.<br><br> 13+sqrt 3 and 7i
vova2212 [387]
So, x = 13, x = √3 and x =7i.

now, recall that for an EVEN radical, there are two possible roots, namely is say √3 is say hmmm some value "a", that means that a*a = √3, however, -a*-a is also √3, therefore, ±√3 are two valid values, and therefore -√3 is another one.

now.... keep in mind that, complex solutions or roots, never come all by their lonesome, their sister is always with them, the conjugate, so, for 7i or namely 0 + 7i, her sister is always around, 0 - 7i, which is the other root.
4 0
3 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
2 years ago
c) The GCF and LCM of two numbers are 2 and 70 respectively. If one of the numbers is 10, find the other number? [3m] solution:
Mice21 [21]

Their LCM is

70

hope it helps you.

8 0
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3 years ago
Which is not true <br><br> 10 + 9 = 7 + 12<br> 10 = 10<br> 10 - 4 = 3 + 3<br><br> 10 = 19 - 11
Natasha2012 [34]

Answer:

10=19-11

Step-by-step explanation:

that answer is false because 19-11 = 8 not 10

3 0
2 years ago
Read 2 more answers
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