At age 77, Eastman suffered from chronic pain due to a spinal condition. On March 14, 1932, after a reportedly jovial meeting with friends, he excused himself to write a brief letter before he 4535

6 0

3 years ago

I have no clue on this one either

taurus [48]

Answer:

Area of the triangle = 210 cm^2 .

Step-by-step explanation:

Perimeter of triangle = 4x + 1 + 3x - 1 + 3x = 70

10x = 70 - 1 + 1 = 70

x = 7.

Area A = 1/2 * base * height

= 1/2 * 3x * (3x - 1)

= 1/2 * 3(7) * (3(7) -1)

= 1/2 * 21 * 20

= 210.

4 0

2 years ago

HELPPPP PLSSS <3 TYYy

olasank [31]

Step-by-step explanation:

On solving denominator →

(16x⁴)^¼ = [(2x)⁴]^¼ = 2x ....eq.1 { 16x⁴ = (2x)⁴ }

Numerator = 2x²

fraction = 2x² / 2x { from eq.1 }

Simplest form = x

3 0

2 years ago

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after subm

True [87]

Answer:

A) cooling constant = 0.0101365

B) $\frac{df}{dt} = k ( 60 - F )$

c) F(t) = 60 + 77.46 $e^{0.0101365t}$

D)137.46 ⁰

Step-by-step explanation:

water temperature = 60⁰F

temperature of Bar after 20 seconds = 120⁰F

temperature of Bar after 60 seconds = 100⁰F

A) Determine the cooling constant K

The newton's law of cooling is given as

= $\frac{df}{dt} = k(60 - F)$

= ∫ $\frac{df}{dt}$ = ∫ k(60 - F)

= ∫ $\frac{df}{60 - F}$ = ∫ kdt

= In (60 -F) = -kt - c

60 - F = $e^{-kt-c}$

60 - F = $C_{1} e^{-kt}$ ( note : $e^{-c}$ is a constant )

after 20 seconds

$C_{1}e^{-k(20)}$ = 60 - 120 = -60

therefore $C_{1} = \frac{-60}{e^{-20k} }$ ------- equation 1

after 60 seconds

$C_{1} e^{-k(60)}$ = 60 - 100 = - 40

therefore $C_{1} = \frac{-40}{e^{-60k} }$ -------- equation 2

solve equation 1 and equation 2 simultaneously

= $\frac{-60}{e^{-20k} }$ = $\frac{-40}{e^{-60k} }$

= 6 $e^{20k}$ = 4 $e^{60k}$

= $\frac{6}{4} e^{40k}$ = In(6/4) = 40k

cooling constant (k) = In(6/4) / 40 = 0.40546 / 40 = 0.0101365

B) what is the differential equation satisfied

substituting the value of k into the newtons law of cooling)

60 - F = $C_{1} e^{0.0101365(t)}$

F(t) = 60 - $C_{1} e^{0.0101365(t)}$

The differential equation that the temperature F(t) of the bar

$\frac{df}{dt} = k ( 60 - F )$

C) The formula for F(t)

t = 20 , F = 120

F(t ) = 60 - $C_{1} e^{0.0101365(t)}$

120 = 60 - $C_{1} e^{0.0101365(t)}$

$C_{1} e^{0.0101365(20)}$ = 60

$C_{1} = 60 * 1.291$ = 77.46

C1 = - 77.46⁰ as the temperature is decreasing

The formula for f(t)

= F(t) = 60 + 77.46 $e^{0.0101365t}$

D) Temperature of the bar at the moment it is submerged

F(0) = 60 + 77.46 $e^{0.01013659(0)}$

F(0) = 60 + 77.46(1)

= 137.46⁰

5 0

4 years ago