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kolbaska11 [484]
3 years ago
14

What is the first step in solving the equation x2 – 16/25 = 0?

Mathematics
2 answers:
Alexeev081 [22]3 years ago
4 0
Add 16/25 to both sides. 

Take the square root of both sides. 



Keith_Richards [23]3 years ago
3 0

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

                     x^2-(16/25)=0 

Step by step solution :<span>Step  1  :</span> 16 Simplify —— 25 <span>Equation at the end of step  1  :</span><span><span> 16 (x2) - —— = 0 25 </span><span> Step  2  :</span></span>Rewriting the whole as an Equivalent Fraction :

<span> 2.1 </span>  Subtracting a fraction from a whole 

Rewrite the whole as a fraction using <span> 25 </span> as the denominator :

<span> x2 x2 • 25 x2 = —— = ——————— 1 25 </span>

<span>Equivalent fraction : </span>The fraction thus generated looks different but has the same value as the whole 

<span>Common denominator : </span>The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

<span> 2.2 </span>      Adding up the two equivalent fractions 
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

<span> x2 • 25 - (16) 25x2 - 16 —————————————— = ————————— 25 25 </span>Trying to factor as a Difference of Squares :

<span> 2.3 </span>     Factoring: <span> 25x2 - 16</span> 

Theory : A difference of two perfect squares, <span> A2 - B2  </span>can be factored into <span> (A+B) • (A-B)

</span>Proof :<span>  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 <span>- AB + AB </span>- B2 = 
        <span> A2 - B2</span>

</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication. 

Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.

Check :  25  is the square of  5 
Check : 16 is the square of 4
Check : <span> x2  </span>is the square of <span> x1 </span>

Factorization is :       (5x + 4)  •  (5x - 4) 

<span>Equation at the end of step  2  :</span> (5x + 4) • (5x - 4) ——————————————————— = 0 25 <span>Step  3  :</span>When a fraction equals zero :<span><span> 3.1 </span>   When a fraction equals zero ...</span>

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the <span>denominator, </span>Tiger multiplys both sides of the equation by the denominator.

Here's how:

(5x+4)•(5x-4) ————————————— • 25 = 0 • 25 25

Now, on the left hand side, the <span> 25 </span> cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (5x+4)  •  (5x-4)  = 0

Theory - Roots of a product :

<span> 3.2 </span>   A product of several terms equals zero.<span> 

 </span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span> 

 </span>We shall now solve each term = 0 separately<span> 

 </span>In other words, we are going to solve as many equations as there are terms in the product<span> 

 </span>Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

<span> 3.3 </span>     Solve  :    5x+4 = 0<span> 

 </span>Subtract  4  from both sides of the equation :<span> 
 </span>                     5x = -4 
Divide both sides of the equation by 5:
                     x = -4/5 = -0.800 

Solving a Single Variable Equation :

<span> 3.4 </span>     Solve  :    5x-4 = 0<span> 

 </span>Add  4  to both sides of the equation :<span> 
 </span>                     5x = 4 
Divide both sides of the equation by 5:
                     x = 4/5 = 0.800 

<span><span> x = 4/5 = 0.800
</span><span> x = -4/5 = -0.800
</span></span>
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Murrr4er [49]

Choice A is the answer which is the point (1,-1)

=========================================

How I got this answer:

Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.

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checking choice A

plug in (x,y) = (1,-1)

y - 2x \le -3

-1 - 2(1) \le -3

-3 \le -3

This is true because -3 is equal to itself. So this is the answer.

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checking choice B

plug in (x,y) = (2,4)

y - 2x \le -3

4 - 2(2) \le -3

0 \le -3

This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.

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checking choice C

plug in (x,y) = (-2,3)

y - 2x \le -3

3 - 2(-2) \le -3

7 \le -3

This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.

-------------

checking choice D

plug in (x,y) = (3,4)

y - 2x \le -3

4 - 2(3) \le -3

-2 \le -3

This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.


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What is 0.6 of 120 really hard and i need an answer for today
ddd [48]
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What is the value of x?. Round to the nearest tenth, if necessary. X = 6 x = 6. 3 x = 7 x = 9. 2.
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Step-by-step explanation:

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qaws [65]

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Show that 1n^ 3 + 2n + 3n ^2 is divisible by 2 and 3 for all positive integers n.
Dmitry_Shevchenko [17]

Prove:

Using mathemetical induction:

P(n) = n^{3}+2n+3n^{2}

for n=1

P(n)  = 1^{3}+2(1)+3(1)^{2} = 6

It is divisible by 2 and 3

Now, for n=k, n > 0

P(k) = k^{3}+2k+3k^{2}

Assuming P(k) is divisible by 2 and 3:

Now, for n=k+1:

P(k+1) = (k+1)^{3}+2(k+1)+3(k+1)^{2}

P(k+1) = k^{3}+3k^{2}+3k+1+2k+2+3k^{2}+6k+3

P(k+1) = P(k)+3(k^{2}+3k+2)

Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also

divisible by 2 and 3.

Hence, by mathematical induction, P(n) = n^{3}+2n+3n^{2} is divisible by 2 and 3 for all positive integer n.

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