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Tanya [424]
3 years ago
11

The beakers shown below have different precisions. label the amount of water in each of the three beakers to the correct number

of significant figures. is it possible for each of the three beakers to the correct number of significant figures

Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

The diagram is attached below

Beaker 1: The measurement line is between 32  and 33. There are 5 lines between 32  and 33. Making line is between third line and fourth line(between 6 and 8). So 32.7

Beaker 2: The measurement line is between 30  and 40. We can clearly see the measurement line at 32. So its 32

Beaker 3: The measurement line is between 32.7  and 32.8. We have 10 small lines inbetween them. Measurement line at 4. So its 32.74

(b) the amount of water in each of 3 beakers are not same because we can measure the difference. But they are close to each other.

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What is the area?<br> 7 m<br> 4 m<br> square meters?
bija089 [108]

Answer:

28 square meters

Step-by-step explanation:

7 x 4= 28

28 square meters

3 0
3 years ago
HELP ME PLZ IN NEED THIS DONE BY 8:10​
Stella [2.4K]

Answer 16.

Step-by-step explanation:

First you replace the xs with 1/2

(√2•1/2+3)/(6•1/2-5)

Then you solve the multiplications

2•1/2=1

6•1/2=3

Now write them in

(√1+3)/(3-5)

Simplify

(√4)(-2)

Root of 4 equals to

2/-2

And the answer is -1

Answer 17:

Replace x with 8

1/2•8^2-(1/4•8+3)

Solve parenthesis, multiplication first

1/2•8^2-(2+3)

1/2•8^2-(5)

1/2•8^2-5

Solve the 8^2 and the 1/2

1/2•64-5

32-5

The answer is 17

3 0
3 years ago
Integral of cosecx dx =log|tanx/2| show that.​
Pie

ANSWER:

Let t = logtan[x/2]

⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx

⇒dt = 1/2 cos² x/2 × cot x/2dx

⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx

⇒dt = 1/2 cosx/2 / sin x/2 dx

⇒dt = 1/sinxdx

⇒dt = cosecxdx

Putting it in the integration we get,

∫cosecx / log tan(x/2)dx

= ∫dt/t

= log∣t∣+c

= log∣logtan x/2∣+c where t = logtan x/2

7 0
3 years ago
Translate into an inequality.
Alex787 [66]
5x-1 is greater than or equal to -11
5 0
3 years ago
Read 2 more answers
Hello there!
otez555 [7]

Answer:

24  Domain: s>=2  or s<=-2

25. 3x^2 +14x +10

26. x^2 -2x+5  

Step-by-step explanation:

24.  Domain is the input or s values

square roots must be greater than or equal to zero

s^2-4 >=0

Add 4 to each side

s^2 >=4

Take the square root

s>=2  or s<=-2


25.  f(g(x))  stick g(x) into f(u) every place you see a u

f(u) = 3u^2 +2u-6

g(x) = x+2

f(g(x) = 3(x+2)^2 +2(x+2) -6

Foil the squared term

       = 3(x^2 +4x+4) +2x+4-6

Distribute

       = 3x^2 +12x+12 +2x+4-6

Combine like terms

   =3x^2 +14x +10


26 f(g(x))  stick g(x) into f(u) every place you see a u

f(u) = u^2+4

g(x) = x-1

f(g(x) = (x-1)^2 +4

Foil the squared term

         = (x^2 -2x+1) +4

         = x^2 -2x+5  

7 0
3 years ago
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