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Brut [27]
3 years ago
10

Will mark brainliest don’t answer false.

Mathematics
1 answer:
andre [41]3 years ago
6 0

The 7 is not a variable because variables are letters. The number seven is a coefficient because they are numbers used to multiply a variable. Example: 6z means 6 times z, and "z" is a variable, so 6 is a coefficient. Example: In ax2 + bx + c, "x" is a variable, and "a" and "b" are coefficients

<em>Thank you! ps can you give me brainly :3</em>

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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})

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2 years ago
Fully explain why an inefficient resource allocation leads an economy to produce at a point inside its production possibilities
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Production possibilities frontier, PPF, is defined as a representation of the point at which a country’s economy is most efficiently producing its goods and services. Efficient production means allocating its resources in the best way possible.
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The points in the PPF shifts outward. The more efficient production is the more the PPF shifts outward. With this in mind, any point found inside the PPF means that there is inefficient resource allocation. If the PPF shifts inward, it means that the efficiency production is regressing and available resources are not fully used to its potential which will lead to a downward turn of the economy. 


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6 0
3 years ago
Jade had the following system of equations to help her with a physics problem, but she only needed to know what the sum of x and
pav-90 [236]
From the second:

2x-y=4

x=(4+y)/2 and applying this to first:

4+y+3y=12

4y+4=12

4y=8

y=2 and since x=(4+y)/2, x=3 so 

x+y=2+3=5
8 0
3 years ago
What is the value of y?​
prohojiy [21]

Answer:

it is 108 because 180-72

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2 years ago
Prove that the product of a nonzero rational number together with an irrational number is an irrational number.
melamori03 [73]
Each time they assume the sum<span> is </span>rational<span>; however, upon rearranging the terms of their equation, they get a contradiction (that an </span>irrational number<span> is equal to a </span>rational number<span>). Since the assumption that the </span>sum of a rational<span> and </span>irrational number<span> is </span>rational<span>leads to a contradiction, the </span>sum<span> must be </span>irrational<span>.</span>
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3 years ago
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