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nadya68 [22]
3 years ago
7

Roberto has a sample of an unknown liquid. The label in the bottle of the sample indicates that the liquid has a mass of 117 gra

ms and a volume of 15 cubic centimeters. What is the liquid in the bottle? gasoline milk water mercury
Mathematics
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

It is Milk

Step-by-step explanation:

The liquid in the bottle has mass given in grams and the mass of that substance is 117 grams.

The volume of the substance is 15 cubic centimeters.

Now, using the formula of density we can estimate which substance could it be.

d=\frac{m}{v}, where 'd', is the density, 'm' is the mass, and 'v' is the volume of the substance.

Plugging the value of mass and volume we get:

d=\frac{117}{15} =7.8g/cm^3

So the density of the substance is 7.8 grams per cubic centimeter.

It cannot be water since the density of water is 1 grams per cubic centimeter.

It cannot be gasoline because its density is 0.77 grams per cubic centimeter.

It cannot be mercury because the density of mercury is 13.56 grams per cubic centimeter.

Since, the density in this case matches the density of thick milk and hence,therefore, the liquid in the bottle is milk.

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jolli1 [7]

Answer:

6.79

Step-by-step explanation:

-Let X denote the total amount of the job.

-The job has to be done in 5 hrs, therefore the portion done every hour is:

Rate=\frac{Total}{Time}\\\\=\frac{X}{5}\\\\=\frac{1}{5}X=0.2X

Therefore, the rate of Bill and Ted working jointly must equal the rate calculated above:

R_{Bill}=\frac{X}{19}\\\\=\frac{1}{19}X\\\\R_{Bill}+R_{Ted}=R_{required}\\\\\frac{1}{19}X+R_{Ted}=\frac{1}{5}X\\\\R_{Ted}=\frac{1}{5}X-\frac{1}{19}X\\\\=\frac{14}{95}X\\\\\therefore \frac{14}{95}X=1\ hr\\\\X=1\div \frac{14}{95}\\\\=6.7857\approx 6.79hrs

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6 0
3 years ago
21. For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2.
dybincka [34]

Answer:

The given lines are perpendicular.

Step-by-step explanation:

In the question, two lines are given as line 1 passes through (1,7) and (5,5). Whereas, line 2 passes through (-1,-3) and (1,1).

It is required to find the slope of given lines and figure out whether they are perpendicular, parallel or neither.

To solve this question, first find the slope of both lines. Check if their product is equal to -1, then they are perpendicular. If the slopes are equal the lines are parallel.

Step 1 of 2

Find the slope of first line.

$$\begin{aligned}m_{1} &=\frac{5-7}{5-1} \\m_{1} &=\frac{-2}{4} \\m_{1} &=-\frac{1}{2}\end{aligned}$$

Step 2 of 2

Find the slope of first line.

$$\begin{aligned}&m_{2}=\frac{1-(-3)}{1-(-1)} \\&m_{2}=\frac{4}{2} \\&m_{2}=2\end{aligned}$$

And

$$\begin{aligned}&m_{1} m_{2}=-\frac{1}{2}(2) \\&m_{1} m_{2}=-1\end{aligned}$$

Since, both slopes are reciprocal of each other.

Therefore, the lines are perpendicular.

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