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kenny6666 [7]
2 years ago
8

What are the first two steps in solving the radical equation below?

Mathematics
1 answer:
Maru [420]2 years ago
6 0

Answer:

B

Step-by-step explanation:

add 7 to get away the 7 on the left hand side

square both side to eliminate square root

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Which transformations will produce similar, but not congruent, figures?
navik [9.2K]
B) Parallelogram JKLM is translated and then dilated
as well as c
C) Parallelogram JKLM is dilated and translated

C is the best answer because is size is changing. In all the others nothing is changing except the position which remains congruent and similar but C does not . The sizes changes which makes them not congruent but still similar.
6 0
3 years ago
Read 2 more answers
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
3 years ago
Please solve this problem and show your work !! Please answer this !!
anastassius [24]

Answer:

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Step-by-step explanation:

Sorry i don't have the step-by-step but that's the answer

4 0
3 years ago
4/5 • 6 = x <br> What is x?
Savatey [412]
X= 4.8 or x= 4 4/5 i believed
8 0
3 years ago
What is the radius of a circle if the area is 78.5 cm^2
Verizon [17]
78.5= (3.14)r^2
25=r^2
5=r
So, the radius of a circle with an area is 78.5 cm^2 is 5 cm^2
3 0
3 years ago
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