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laiz [17]
1 year ago
5

Find the quadratic equation with integral coefficients having roots 6 and 2.

Mathematics
1 answer:
ludmilkaskok [199]1 year ago
4 0

The quadratic equation that has a root of 6 and 2 is x^2 - 8x + 12 = 0

<h3>How to determine the quadratic equation?</h3>

The roots are given as:

x = 6 and x = 2

Rewrite as:

x - 6 = 0 and x - 2 = 0

Multiply both equations

(x - 6)(x - 2) = 0 * 0

Evaluate the product

x^2 - 6x - 2x + 12 = 0

Evaluate the difference

x^2 - 8x + 12 = 0

Hence, the equation that has a root of 6 and 2 is x^2 - 8x + 12 = 0

Read more about quadratic equations at:

brainly.com/question/10449635

#SPJ1

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1)Circle 1 is centered at (−4, 5)(−4, 5) and has a radius of 2 centimeters. Circle 2 is centered at (2, 1)(2, 1) and has a radiu
galina1969 [7]

Figures of same shape and size are similar .Two circles  C1&C2 will be similar.

Circle 1 has a center of (-4,5) and circle 2 has a center of (2,1) .The x of the center is having the translation x+6 and the y  is having a translation of y-4.The center of the circle is dilated by 3 units.

The circles are similar because you can translate Circle 1 using the transformation rule (x+6,y-4 ) and then dilate it using a scale factor of 3.

2) Area of sector = \pi.r^{2}\alpha÷360.

Where α is the angle made at center.

Area of given sector= π(12)(12)(60)÷360 =24π.

8 0
3 years ago
Pls help i am on a timer
stich3 [128]

Answer:

x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}

Step-by-step explanation:

Given;

x² - 2x - 1 = 0

Solve by completing the square method;

⇒ take the constant to the right hand side of the equation.

x² - 2x = 1

⇒ take half of coefficient of x = ¹/₂ x -2 = -1

⇒ square half of coefficient of x and add it to the both sides of the equation

x^2 +  (-1)^2 = 1 + (-1)^2

(x-1)^2 = 1 + 1\\\\(x-1)^2 = 2\\\\

⇒ take the square root of both sides;

x-1 = +/- \ \ \sqrt{2} \\\\x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}

Therefore, option B is the right solution.

3 0
3 years ago
20. What is the equation of the line that passes through the points (–2, 2) and (0, 5)? A. y = x + 5 B. y = x – 5 C. y = –3/2x +
Vanyuwa [196]

y=mx+b

first find the y intercept or b or x=0

b=5

now find the slope

y²-y¹/x²-x¹--> 5-2/0-(-2)

m=3/2

the equation is D. y=3/2x+5

3 0
3 years ago
The decimal number 11.8 can be read as "eleven and eight tens."<br><br><br> True or False
Semmy [17]
I mean it could be read that way but that's the way most people read fractions. 11 and 8/10. Most people I know read decimals like that as 11 point 8. So yes I suppose it's true; it's not wrong, but most people don't say it like that. And I don't know how they taught you to read it in your math course.
3 0
3 years ago
Read 2 more answers
Determine if b is a linear combination of a1 a2, and a3. a1 = [ 1 -2 0 ], a2 = [ 0 1 3 ], a3 = [ 6 -6 18 ], b = [ 2 -2 6 ] Choos
marysya [2.9K]

Answer: Vector b is not a linear combination

Step-by-step explanation:

First of all we put the vectors in terms of different variables, such as:

a1(1,-2,0)=(a,-2a,0);

a2(0,1,3)=(0,b,3b);

a3(6,-6,18)=(6c,-6c,18c);

To know that a vector is a linear combination we need to express it like a sum of other different vectors.

(2,-2,6)=(a,-2a,0)+(0,b,3b)+(6c,-6c,18c)

(2,-2,6)=(a+0+6c,-2a+b-6c,0+3b+18c)

We express this sum like a system of equations.

a+6c=2

-2a+b-6c=-2

3b+18c=6

We solve this system of equations and we can note that the system don't have a solution, so the vector b is not a linear combination of a1, a2, and a3.

8 0
3 years ago
Read 2 more answers
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