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Reptile [31]
2 years ago
15

22 (x+6)(x+6) ++ 3x3x ++ 44 ==

Mathematics
1 answer:
creativ13 [48]2 years ago
4 0
I’m pretty sure the answer is 31x
2
+264x+836
You might be interested in
Two sides of a triangle are 12 and 15. Which inequality represents the solution to finding x, the third side.
Vikentia [17]

Answer:

the other side would be 15

Step-by-step explanation:

because 15 is bigger than 12 and the 12 is the bottom line and then the other line going up is 15 so it's equal.

4 0
2 years ago
find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0​
Citrus2011 [14]

Answer:

Radius: r =\frac{\sqrt {21}}{6}

Center = (-\frac{3}{2}, -\frac{2}{3})

Step-by-step explanation:

Given

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Solving (a): The radius of the circle

First, we express the equation as:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

So, we have:

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Divide through by 9

x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0

Rewrite as:

x^2  + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}

Group the expression into 2

[x^2  + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

Next, we complete the square on each group.

For [x^2  + 3x]

1: Divide the coefficient\ of\ x\ by\ 2

2: Take the square\ of\ the\ division

3: Add this square\ to\ both\ sides\ of\ the\ equation.

So, we have:

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

[x^2  + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Factorize

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Apply the same to y

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}

Add the fractions

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

Recall that:

(x - h)^2 + (y - k)^2 = r^2

By comparison:

r^2 =\frac{7}{12}

Take square roots of both sides

r =\sqrt{\frac{7}{12}}

Split

r =\frac{\sqrt 7}{\sqrt 12}

Rationalize

r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}

r =\frac{\sqrt {84}}{12}

r =\frac{\sqrt {4*21}}{12}

r =\frac{2\sqrt {21}}{12}

r =\frac{\sqrt {21}}{6}

Solving (b): The center

Recall that:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

From:

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

-h = \frac{3}{2} and -k = \frac{2}{3}

Solve for h and k

h = -\frac{3}{2} and k = -\frac{2}{3}

Hence, the center is:

Center = (-\frac{3}{2}, -\frac{2}{3})

6 0
2 years ago
The annual total price is approximately $11,600 for Texas residents. If a student was atiending for 4 years, and received one sc
MatroZZZ [7]
Hbuggguvhvuvhvhvyvhvhvuvjvjvhvhvhvyvgvgvg g g h g h g g g g g. Gvgvhvh
3 0
2 years ago
Jack got the expression 7x+1 and then wrote his answer as 1+7x . is his answer an equivalent expression? how do you know?
Solnce55 [7]
True. According to the Commutative Property <span>of addition, the numbers could be put in any order and still result in the same answer. </span>



3 0
3 years ago
Jose can paint an entire house in seven hours and brandon can paint the same house in eight hours. Write an equation that can be
guajiro [1.7K]

Answer:

t/8 + t/7 = 1

Step-by-step explanation:

Given in the question that,

time require for Jose to paint the house = 7 hours

time require for Brandon to paint the house = 8 hours

Suppose t means Full house painted.

<h3>To solve the question we have to figure out how much each of them can paint in ONE hour.</h3>

7 hours----t

1 hour ---- t/7

8 hours----t

1 hour ---- t/8

<h3>Equation</h3>

t/8 + t/7 = 1 (in one hour)

(7t + 8t)/8(7) = 1

15t/56 = 1

15t = 56

t = 56/15

t = 3.73 hours

5 0
3 years ago
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