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3 years ago

Solve each equation.1. 6k – 3 = 2k + 13

Mathematics

1 answer:

NARA [144]3 years ago

4 0

Answer:

4k=16

k=4

Step-by-step explanation:

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A pan of warm water (46∘ Celsius) was put in a refrigerator. After 15 minutes, the water's temperature was 27∘ C; 15 minutes aft

Cloud [144]

Answer:

The refrigerator was at around 13 ∘C

Step-by-step explanation:

Newton's Law of Cooling:

The rate of change of a body temperature (amount of heat loss/time of loss) is directly proportional to the difference between its own temperature and the surroundings.

$\frac{T(t2)-T(t1)}{t2-t1} = -h (T(t2) - Tenv)$

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

$\frac{T(t2)-T(t1)}{t2-t1}$ ⇒ rate of change of he temperature

-h ⇒ constant of proportionality (negative because the temperature is decreasing inside the refrigerator)

We have 3 points:

time (minutes) - Temperature (∘ C)

0 (when the pan was put in the refrigerator) - 46

15 (after 15 minutes) - 27

30 (15 minutes after the first 15 minutes) - 19

$\frac{27 - 46}{15 - 0}$ = -h (27 - Tenv)

$\frac{19 - 27}{15 - 0}$ = -h (19 - Tenv)

Now we have a system of two equations and two variables

$\frac{-19}{15} = -h (27 - Tenv)\\ \frac{19}{15} = h (27 - Tenv)\\ \frac{19}{15(27 - Tenv)} = h$

$\frac{19 - 27}{15 - 0} = -h (19 - Tenv)\\\frac{-8}{15} = -h (19 - Tenv)\\\frac{8}{15} = h (19 - Tenv)\\\frac{8}{15} = \frac{19}{15(27 - Tenv)} (19 - Tenv)\\8(27 - Tenv) = 19 (19 - Tenv)\\216 - 8 Tenv = 361 - 19 Tenv\\19 Tenv - 8 Tenv = 361 - 216\\11 Tenv = 145 \\Tenv = 145/ 11\\Tenv = 13. 18$

The refrigerator was at around 13 ∘C

6 0

3 years ago

Help I will mark you brainly

Ludmilka [50]

Answer:

Hi there!!

The answer would be x=20 and y= 10 root 3 or on decimal it's 17.32.

explanation look in picture, alright. 

I hope it will help u....

5 0

3 years ago

Read 2 more answers

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$