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2 years ago

N^2+6n+c Find the value of c that completes the square

Mathematics

1 answer:

Anon25 [30]2 years ago

6 0

Answer:

The value of ”c” can by found by this formula: (b/2)^2. In this equation the ”b” is 6. So, (6/3)^2=3^2=9.

The value of “c” that completes the square is 9.

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Answer:

True

False

Step-by-step explanation:

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Let G=gcd(a,b). Since r=a-bq and G divides a and G divides b, then G divides r. Now, G divides a and G divides r, hence G divides gcd(b,r).

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x divides y and y divides x implies that |x|=|y|. The GCD's are nonnegative, therefore G=gcd(b,r).

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101 is prime, but this may be false in general. For example, consider N=13*11=143. N is not prime, and n is not divisible by 2,3,5, or 7.

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3 years ago

A pool gained 10 gallons of water on Friday due to rain. How could the owner of the pool bring the water level back to its previ

Len [333]

Answer

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Step-by-step explanation:

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3 years ago

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hey download photomath it works great u take a picture of your math problem and itll give u the steps n the answer

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3 years ago

Ella’s Cafe uses 3/10 of a bag of coffee every day. For how many days will 2 7/10 bags of coffee last? Simplify your answer and

Vilka [71]

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3 years ago

Read 2 more answers

Solving Quadratic Equations with Complex SolutionsThe discriminant: D=b^2-4acQuestion:5x^2-2x+1=0

fomenos

• The value of the discriminant ,D= -16

• The solution to the quadratic equation is

$x=\frac{1+2i}{5}\text{ or }\frac{1-2i}{5}$

Step - by - Step Explanation

What to find?

• The discriminant d= b² - 4ac

• The solution to the quadratic equation.

Given:

5x² - 2x + 1=0

Comparing the given equation with the general form of the quadratic equation ax² + bx + c=0

a=5 b=-2 and c=1

Uisng the quadratic formula to solve;

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

The discriminant D=b² - 4ac

Substitute the values into the discriminant formula and simplify.

D = (-2)² - 4(5)(1)

D = 4 - 20

D = -16

We can now proceed to find the solution of the quadratic equation by substituting into the quadratic formula;

$x=\frac{-(-2)\pm\sqrt[]{-16}}{2(5)}$

Note that:

√-1 = i

$x=\frac{2\pm\sqrt[]{16\times-1}}{10}$

$x=\frac{2\pm\sqrt[]{16}\times\sqrt[]{-1}}{10}$ $x=\frac{2\pm4i}{10}$ $x=\frac{2}{10}\pm\frac{4i}{10}$ $x=\frac{1}{5}\pm\frac{2}{5}i$ $x=\frac{1\pm2i}{5}$

That is;

$\text{Either x=}\frac{1+2i}{5}\text{ or x=}\frac{1-2i}{5}$

7 0

1 year ago