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vitfil [10]
3 years ago
6

Which set of values could be from a direct proportion

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

"A"

Step-by-step explanation:

I got it correct

mr Goodwill [35]3 years ago
3 0

Answer:

First table

Step-by-step explanation:

We have to find the set of value which could be from a direct proportion.

We know that

Direct proportion :

When x is directly proportional to y

x\propto y

x=ky

Where k=Proportionality constant

k=\frac{x}{y}

K remain constant when x and y are in direct proportion

From first table

k=\frac{\frac{1}{2}}{6}=\frac{1}{12}

k=\frac{1}{12}

k=\frac{2}{24}=\frac{1}{12}

K=\frac{3}{36}=\frac{1}{12}

When x and y are both varies then ratio of x and y remain constant.Hence, it is in direct proportion.

From second table

k=\frac{0}{0}= not ]define

k=\frac{1}{1}=1

It is not direct proportion because k does not remain constant.

From third table

k=\frac{0}{2}=0

k=\frac{2}{4}=\frac{1}{2}

It is not direct proportion because k does not remain constant.

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Step-by-step explanation:

Let a, b, and c represent the earnings of Alan, Bob, and Charles. The problem statement tells us ...

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Adding the first and third equations, we get ...

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Alan earned $110.

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<em>Check</em>

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A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
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\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

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