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polet [3.4K]
3 years ago
11

One number is 6 more than anouther number. the sum of their squares is 90

Mathematics
1 answer:
Flura [38]3 years ago
6 0
Let x = greater number
      y = smaller  number

(1) x = y + 6

(1) y = x-6

(2) x^{2} + y^{2} = 90
We'll substitute y in (1) to (2)
(2) x^{2} +  (x-6)^{2} = 90
     x^{2} +  (x^{2} - 12x + 36) = 90
     x^2 + x^2 -12x + 36 - 90 = 0
     2x^2 - 12x -54 = 0

     2(x^2 - 6x - 27) = 0
     2(x - 9)(x + 3) = 0
 x - 9 = 0        or      x + 3 = 0
      x = 9                        x = -3
and
 y = x - 6                  y = x - 6
 y = 9 - 6                  y = -3 - 6
 y = 3                        y = -9

Therefore, the two numbers can be 9 and 3 or -3 and -9.

      
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m=4

Step-by-step explanation:

Since we know that quadrilateral ABCD is similar to QRST, we know that the side lengths will be proportional to one another. As such, we should take a ratio to determine the side length of m.

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3 years ago
Find the standard equation of the circle with center (5, -3
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Answer:

○ C

Explanation:

Accourding to one of the circle equations, \displaystyle (x - h)^2 + (y - k)^2 = r^2,the centre of the circle is represented by \displaystyle (h, k).Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:

Pythagorean Theorem

\displaystyle a^2 + b^2 = c^2 \\ \\ 6^2 + 8^2 = r^2 \hookrightarrow 36 + 64 = r^2 \hookrightarrow \sqrt{100} = \sqrt{r^2} \\ \\ \boxed{10 = r}

Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.

Distanse Equation

\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-5 - 3]^2 + [3 + 3]^2} = r \hookrightarrow \sqrt{[-8]^2 + 6^2} = r \hookrightarrow \sqrt{64 + 36} = r; \sqrt{100} = r \\ \\ \boxed{10 = r}

Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.

I am joyous to assist you at any time.

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During a game, players can win and lose counters. At the start of the game
Mademuasel [1]
The first step is to calculate what fraction of the counters that Rob has.

Add all the portions together to find the total: 5+6+7= 18
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Next, do the same for the end ratio:
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By comparing his beginning portion (0.278) to his end portion (0.292) it is clear that he has a greater portion of the total counters. Given that the problem states that the number of counters stays the same, Rob must have more counters than he started with.
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