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lana [24]
3 years ago
13

Which ordered pair is a solution of the equation? 3x-y=13

Mathematics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

(\frac{13}{3},0) and

Step-by-step explanation:

3x-y=13 \\ \\ -y=-3x+13 \\ \\ y =3x-13

Any point on this line will do.

For this example, let's find the x and y intercepts by plugging in 0 for x and y.

You'll get (\frac{13}{3},0) and (0,-13)(0,-13)

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These two rectangles are similar which is a correct proportion for corresponding sides
hammer [34]

to make the proportion we make the ratio of the similar sides

For example

4/8 = 12/x

or we can write it as 4/12 = 8/x

or 12/x = 4/8

So last option is correct

4 0
3 years ago
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1. Pedro is a good student.<br> a.<br> b.
ICE Princess25 [194]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
(x-6)² + (x + 5)² = 16
Phoenix [80]

Answer:

answer A   : (2 -5)

Step-by-step explanation:

the midipoint P,Q is the center A of this circle (because PQ is a diameter)

use : (x-6)²+(x+5)² = 16 you have : A(6 , -5) now  let Q(x,y) :

 (x+10)/2 = 6

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6 0
3 years ago
The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?
pashok25 [27]

Answer:

Yes, it is true that x^2+3x-1  is a factor of x^4+3x^3-2x^2-3x+1.

Step-by-step explanation:

Let us try to factorize x^4+3x^3-2x^2-3x+1

x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3

Let us try to make a whole square of the given terms:

\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\

--------------

Formula used above:

a^{2} -2 \times a \times b +b^{2}  = (a-b)^2

In the above equation, we had a = x, b = 1.

--------------

Further solving the above equation, taking 3x common out of 3x^3-3x

\Rightarrow (x^2-1)^2+3x(x^2-1)\\

Taking (x^{2} -1) common out of the above term:

\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)

So, the two factors are (x^2-1)\ and\ (x^2+3x-1).

\therefore The statement that x^2+3x-1  is a factor of x^4+3x^3-2x^2-3x+1 is <em>True.</em>

7 0
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4.76 to one decimal place
stellarik [79]
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3 0
3 years ago
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