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Oksanka [162]
3 years ago
10

Given the equation 2 Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

Mathematics
2 answers:
leonid [27]3 years ago
8 0
<span>2 square root (x - 5) = 2. 2^2 (x -5) = 2^2. 4(x - 5) = 4. 4x - 20 = 4. 4x = 24. x = 6. This solution is not extraneous, because extraneous solutions emerge from solving the problem but are not actually valid solutions for the initial problem. With rounding, this solution is valid for the initial problem.</span>
mylen [45]3 years ago
8 0

Answer:

x = 6, solution is not extraneous

Step-by-step explanation:

Given the equation 2 Square root of x minus 5 = 2

2\sqrt{x-5} =2. solve for x

Divide by 2 on both sides

\sqrt{x-5} =1

Take square on both sides to remove square root

x-5= 1

add 5 on both sides

x= 6

Now we plug in x=6 in our original equation

2\sqrt{6-5} =2

2 = 2 is true

So x= 6 satisfies our given equation

X=6 is not an extraneous solution

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The value of integration of y=16-x^{2} from x=-1 to x=1 is 94/3.

Given the equation y=16-x^{2} and the limit of the integral be x=-1,x=1.

We are required to find the value of integration of y=16-x^{2} from x=-1 to x=1.

Equation is relationship between two or more variables that are expressed in equal to form.Equation of two variables look like ax+by=c.It may be linear equation, quadratic equation, or many more depending on the power of variable.

Integration is basically opposite of differentiation.

y=16-x^{2}

Find the integration of 16-x^{2}.

=16x-x^{3}/3

Now find the value of integration from x=-1 to x=1.

=16(1)-(1)^{3}/3-16(-1)-(-1)^{3}/3

=16(1)-1/3+16-1/3

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Hence the value of integration of y=16-x^{2} from x=-1 to x=1 is 94/3.

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4 0
1 year ago
PLZ ANSWER IF U KNOW THE ANSWER​
creativ13 [48]

Answer: its b

Step-by-step explanation:

3 0
3 years ago
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Graph the first six terms of a sequence where a1=5 and r=1.25
emmainna [20.7K]
The sequence above is geometric progression. 
The nth term of such sequence is given by;

Tn = ar∧(n-1),
Where a⇒first term and
            r⇒common ratio

So, 1st term = 5×1.25∧(1-1) = 5×1.25∧0 =5 
      2nd term = 5×1.25∧(2-1) = 5×1.25 = 6.25
      3rd term = 5×1.25∧(3-1) = 5×1.25² = 7.8125
       4th term = 5×1.25∧(4-1) =5×1.25³ = 9.765625
       5th term = 5×1.25∧(5-1) = 5×1.25∧4 = 12.20703125
       6th term = 5×1.25∧(6-1) = 5×1.25∧5 = 15.25878909
7 0
3 years ago
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