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pshichka [43]
4 years ago
8

When a substance freezes if changes from the liquid phase to the solid phase because of this

Mathematics
2 answers:
Jet001 [13]4 years ago
5 0

well the name of the phase for that is called freezing. I think that's we have to look for, maybe not, but I hope I helped

Contact [7]4 years ago
4 0
Chrystal the other part nation but without the na but with an iza
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*) Eric works at the deli on weekends to earn extra money. He makes $10 per hour making
algol13

Answer:

A=\dfrac{1}{2}(10s+14d)

Step-by-step explanation:

Given that,

He makes $10 per hour making  sandwiches, and $14 per hour delivering orders.

Let there are s sandwiches and d is delivering orders.

Eric puts half of his total earnings in a  savings account for college.

The expression that represents the total amount of money Eric saves for working s hours making sandwiches and d hours delivering orders.

A=\dfrac{1}{2}(10s+14d)

Hence, this is the required solution.

8 0
3 years ago
1. Write the equation c = loga(b) in exponential form.
Flauer [41]

Answer:

Step-by-step explanRemember: logax=b⇒ab=x

in which "a" is the base of the log. 

log4 (1/16) = 4^-2 = 1/4^2 = 1/16

4 to the power of -2 is equal to 1/16.

To answer this question, the equation in exponential form of "log4 (1/16) = -2" is " 4^-2 " (4 to the power of -2). :D

plz mark me brainliest

7 0
3 years ago
Which of the following is 1 2 over 4 in its simplest form?
Katen [24]

Answer:

its 3

Step-by-step explanation:

3 is your answer I hope this help

6 0
3 years ago
Read 2 more answers
I’m really dum man and help me pls
Llana [10]

Answer:

18 /25

Step-by-step explanation:

Multiply

3/5  × 6/5  = 18/25

3 /5  × 6/5  = ( 3 × 6 ) over ( 5 × 5 )

= 18/25

6 0
3 years ago
Read 2 more answers
A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position
pantera1 [17]

Answer:

233.48s  

3.84 min

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.

We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:

y=A cos(\omega t+\phi)+C

where:

A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

\omega= angular speed.

in this case:

\omega=\frac{2\pi}{T}

where T is the period, in this case 6 min or

6min(\frac{60s}{1min})=360s

so:

\omega=\frac{2\pi}{360}

\omega = \frac{\pi}{160}

and

\phi= phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

y=-20 cos(\frac{\pi}{180}t)+22

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

-20 cos(\frac{\pi}{180}t)+22>13

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

-20 cos(\frac{\pi}{180}t)+22=13

and solve it:

-20 cos(\frac{\pi}{180}t)=13-22

-20 cos(\frac{\pi}{180}t)=-9

cos(\frac{\pi}{180}t)=\frac{9}{20}

and we can take the inverse of cos to get:

\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})

which yields two possible answers: (see attached picture)

so

\frac{\pi}{180}t=1.104 or \frac{\pi}{180}t=5.179

so we can solve the two equations. Let's start with the first one:

\frac{\pi}{180}t=1.104

t =1.104(\frac{180}{\pi})

t=63.25s

and the second one:

\frac{\pi}{180}t=5.179

t=5.179(\frac{180}{\pi})

t=296.73s

so now we can build our possible intervals we can use to test the inequality:

[0, 63.25]  for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25]  for a test value of 1

-20 cos(\frac{\pi}{180}(1))+22>13

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

-20 cos(\frac{\pi}{180}(70))+22>13

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

-20 cos(\frac{\pi}{180}(300))+22>13

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

233.48(\frac{1min}{60s})=3.84 min

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

3 0
3 years ago
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