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nlexa [21]
4 years ago
12

What is six copies of the sum of nine-fifths and three

Mathematics
1 answer:
Ira Lisetskai [31]4 years ago
7 0
9/5+3/1


24/5 24/5 24/5 24/5 24/5 24/5
Hope this helps! :D
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Find the equation of the line, where P is a point on the graph of the line. m=2/3 , P(3,5)
jeka57 [31]

Answer:

y = 2/3x + 3

Step-by-step explanation:

y - 5 = 2/3(x - 3)

y - 5 = 2/3x - 2

y = 2/3x - 2 + 5

y = 2/3x + 3

3 0
3 years ago
Arrange the hyperbolas in increasing order of the horizontal widths of their asymptote rectangles.
almond37 [142]

The  increasing order of the horizontal widths of their asymptote rectangles is dependent on the values gotten from  y = ± x.

<h3>What is a Hyperbola?</h3>

This is defined as a two-branched open curve formed by the intersection of a plane perpendicular to the bases of a double cone.

The rectangular hyperbola has two asymptotes which are defined as y = ± x in this scenario.

Read more about Hyperbola here brainly.com/question/3351710

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2 years ago
If a number is decreased by 8, the result is -21. What is the
wariber [46]

Answer:

answer = -13

Step-by-step explanation:

number = x

x - 8 = -21

x = -21 + 8

  = -13

Recheck :

num = - 13

decreased by 8 = - 13 -8 = -21

4 0
3 years ago
(5x+2)[3(x+14)]<br><br> what is the value of x
Julli [10]

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5 0
3 years ago
Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}
miskamm [114]

Answer:

-3+i\sqrt{3} , 1+\sqrt{3}

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

\alpha = x+iy\\\beta = x-iy

since they are conjugates

\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}

\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}

Imaginary part of the above =0

i.e. \sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1

So the value of alpha = -3+i\sqrt{3} , 1+\sqrt{3}

3 0
3 years ago
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