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4 years ago

5

Johnny earned a total of $15 for 3

1 answer:

solniwko [45]4 years ago

4 0

Answer:

5 dollars a hour. ,

Step-by-step explanation:

count by fives

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3 years ago

Prove that:<br><br>cos20°cos40°cos80°=1/8

Naya [18.7K]

Answer:

see explanation

Step-by-step explanation:

Using the double angle identity for sine

sin2x = 2sinxcosx

Consider left side

cos20°cos40°cos80°

= $\frac{1}{2sin20}$ (2sin20°cos20°)cos40°cos80°

= $\frac{1}{4sin20}$ (2sin40°cos40°)cos80°

= $\frac{1}{4sin20}$ (sin80°cos80° )

= $\frac{1}{8sin20}$ (2sin80°cos80° )

= $\frac{1}{8sin20}$ . sin160°

= $\frac{1}{8sin20}$ . sin(180 - 20)°

= $\frac{1}{8sin20}$ . sin20°

= $\frac{1}{8}$ = right side , thus proven

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3 years ago

Read 2 more answers

PLEASE ANSWER<br><br> What is an outlier in math?

neonofarm [45]

A value that "lies outside" (is much smaller or larger than) most of the other values in a set of data.

For example in the scores 25,29,3,32,85,33,27,28 both 3 and 85 are "outliers".

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4 years ago

Send help please, i’m really confused!

GarryVolchara [31]

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2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago