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hoa [83]
2 years ago
10

A triangle with sides 11m , 13m and 18m is a right triangle. ​ A True B False

Mathematics
1 answer:
Julli [10]2 years ago
5 0

\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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(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference betw
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Answer:

a. \frac{dT}{dt}=k(T-Tm); T(0)=190

b. C_{0}=122

c. k=-0.00259

d. t=153.39838\\ minutos

Step-by-step explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

Tm=68

\frac{dT}{dt}=k(T-Tm); T(0)=190

b. The differential equation obtained is a differential equation of separable variables:

\frac{dT}{T-Tm}=kdt\\\\\int {\frac{dT}{T-Tm}}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_{0}e^{kt}+Tm=C_{0}e^{kt}+68\\\\T(0)=C_{0}e^{k(0)}+68=190\\\\C_{0}=122

c. After 33 minutes of serving the coffee has cooled to 180°:

T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259

d.

150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838\\\\

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