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valkas [14]
3 years ago
5

Metal Melting Point (°C) Boiling Point (°C) Density in Solid State (g/cm3) Copper 1,083 2,567 8.3–9.0 Lead 327.3 1,750 11.3 Gold

1,063 2,660 19.3 A scientist has a sample of each of the metals listed in the table above. All three samples have the same mass. Of the three samples, the sample of _______ will change from a liquid to a gas at the highest temperature, and the sample of _______ has the greatest volume. A. gold; copper B. copper; gold C. copper; lead D. gold; gold
Mathematics
2 answers:
WINSTONCH [101]3 years ago
8 0
Based from the given data of the metals' melting point, boiling point, and density, the answer is A. The sample of gold will change from liquid to a gas at the highest temperature, and the sample of copper has the greatest volume. Gold has the highest boiling point of 2660 while copper has the lowest density which can correspond to a large volume knowing that the density is mass over volume. 
Darya [45]3 years ago
3 0

Answer:A

Step-by-step explanation:

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0.01 × 0.001<br>answer This​
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Answer:

0.00001

Step-by-step explanation:

6 0
2 years ago
What is 48,148 in expanded form?
irina1246 [14]

Answer:

4 ten thousands, 8 thousands, 1 hundred, 4 tens, 8 ones.

Step-by-step explanation:

6 0
3 years ago
Solve for x. enter your answer in interval notation using grouping symbols. x^2+15x&lt;-36
Oliga [24]
X²+15x+36<0

at first solve quadratic equation

D=b²-4ac= 225-4*1*36= 81

x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3

we can write x²+15x+36<0 as (x+12)(x+3)<0

(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
 1 case)      x+12<0, and x+3>0,                                            
                       x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set

or second case)  x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3)  they are crossing , so (-12, -3)  is a solution of this inequality


3 0
3 years ago
Determine an exact value for sin _<br> 3π/4-tan 5π/6
iris [78.8K]

Answer:

\frac{3\sqrt{2}+2\sqrt{3}}{6} or \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}

Step-by-step explanation:

\sin\frac{3\pi}{4}-\tan\frac{5\pi}{6}\\\\\sin\frac{3\pi}{4}-\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}\\ \\\frac{\sqrt{2}}{2}-\frac{\frac{1}{2} }{-\frac{\sqrt{3}}{2}}\\ \\\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{3}}\\ \\\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}\\ \\\frac{3\sqrt{2}}{6}+\frac{2\sqrt{3}}{6}\\ \\\frac{3\sqrt{2}+2\sqrt{3}}{6}

6 0
2 years ago
Solve for x in the figure below<br> 5x+18<br> 7x-14
daser333 [38]

Answer:

16 =x

Step-by-step explanation:

The angles are corresponding angles and since the lines are parallel, they are equal

5x+18 =7x-14

Subtract 5x from each side

5x+18 -5x=7x-5x-14

18 = 2x-14

Add 14 to each side

18+14 = 2x-14+14

32 = 2x

Divide by 2

32/2 = 2x/2

16 =x

8 0
3 years ago
Read 2 more answers
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