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3 years ago

Please help me with this problem!

Mathematics

1 answer:

zimovet [89]3 years ago

6 0

2 x $1150 + $1049.34 = $3,349.34.

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Answer

According to newton law of cooling

$\dfrac{d T}{dt} = k (T - T_a)$

and

$T = Ce^{kt} + T_a$

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

$T(0) = Ce^{0} + T_a$

10 = C + 100

C = -90

now,

$12 = -90e^k + 100$

$e^k = \dfrac{88}{90}$

$k = ln{\dfrac{88}{90}}$

at time equal to t

T(t) = 70

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$70 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{3}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})$

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$98 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{45}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})$

t = 390 s

hence, after 390 s the temperature of the body will be 98°C

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You must show your staterment and reason T-chart to get full credit

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Step-by-step explanation:

RS bisects PQ at T. PQ bisects RS at T.

<u>prove</u>: Triangle PTS congruent Triangle QTR

<h3>Statements..............................Reasons</h3>

(1) RS bisects PQ at T.................(1) Given

(2) PQ bisects RS at T.................(2) Given

(3) angle PST ≅ angle QRT........(3) supplements of congruent angles

(4) TS ≅ TR......................(4) Converse of Base Angles Theorem

(5) angle PTS ≅ angle QTR.........(5) Subtraction Property of Equality

(6) Triangle PTS ≅ Triangle QTR...(6) Angle side angle

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2 years ago