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Brums [2.3K]
2 years ago
13

1. Is (0,9) a solution? y = -3x + 5

Mathematics
2 answers:
nikdorinn [45]2 years ago
6 0

1. Is (0,9) a solution?

y = -3x + 5

\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}

<h3>Solution⤵️</h3>

Let us observe the given things;

  • <u>An</u><u> </u><u>equation</u><u> </u><u>is</u><u> </u><u>given</u><u> </u><u>with</u><u> </u><u>variables</u><u> </u><u>x</u><u> </u><u>and</u><u> </u><u>y</u>
  • <u>The</u><u> </u><u>values</u><u> </u><u>of</u><u> </u><u>x</u><u> </u><u>and</u><u> </u><u>y</u><u> </u><u>are</u><u> </u><u>given</u>
  • <u>We</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>find</u><u> </u><u>whether</u><u> </u><u>they</u><u> </u><u>are</u><u> </u><u>the</u><u> </u><u>correct</u><u> </u><u>values</u><u> </u><u>or</u><u> </u><u>not</u>
  • <u>Let</u><u> </u><u>us</u><u> </u><u>verify</u><u> </u><u>if</u><u> </u><u>they</u><u> </u><u>are</u><u> </u><u>a</u><u> </u><u>solution</u><u> </u><u>or</u><u> </u><u>not</u><u>!</u><u>~</u>

\sf ⟾ \:  \red{y =  - 3x + 5}

Plug in the x and y values I.e. x=0 & y=9 respectively

\sf⟾  \: \orange{9 =  - 3 \times 0 + 5}

Multiply the numbers

\sf ⟾ \:  \green{9 = 0 + 5}

Add the numbers

\sf⟾ \:  \blue{9 = 5}

Check the equality

\sf⟾ \:  \purple{9 \cancel = 5}

They are not equal!

<em>Since</em><em>,</em><em> </em><em>They</em><em> </em><em>are</em><em> </em><em>not</em><em> </em><em>equal</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>conclude</em><em> </em><em>that</em><em> </em><em>the</em><em> </em><em>given</em><em> </em><em>coordinates</em><em> </em><em>are</em><em> </em><em>not</em><em> </em><em>a</em><em> </em><em>solution</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>given</em><em> </em><em>equation</em><em>!</em><em>!</em><em>~</em>

irinina [24]2 years ago
4 0

Answer:

no

Step-by-step explanation:

to determine if (0, 9 ) is a solution substitute x = 0 into the equation and if the result is 9 then it is a solution

y = - 3(0) + 5 = 0 + 5 = 5 ≠ 9

then (0, 9 ) is not a solution to the equation

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A point in the figure is selected at random. Find the probability that the point will be in the part that is NOT shaded.
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Answer:

B

Step-by-step explanation:

So the question really is what is the probability that the point is in the square? You cannot pick something surrounding the circles because you do not know anything about that region.

This is solved by comparing areas. There are 2 full circles there. 1 circle has an area of pi*r^2

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Let r = 3   This is a completely random choice. No matter what number you choose for r, the answer will come out the same. When you get along a little further in math, you will find that you can just use letters.

Total shaded area = 2 * 3.14 * 3^2

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Answer

% = (area of square) * 100% / Total area of both regions

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Obviously the answer I get is not offered. The closest answer is B so I will choose that.

==========================

This is how this question would be done without using any value for r.

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Area of square to total (as a %) = 4/10.28  * 100 = 39% which gives the same answer.

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