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Shalnov [3]
3 years ago
10

Gil is reading a book that has 276 pages.he already read some of it last week. He plans to read 20 pages tomorrow. By then he wi

ll be 2/3 of the way through the book. How many pages did Gil read last week?
Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0
Last week Gill read 164 of the 276 pages of the book.
                                                                   hope this helped :)








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CAN SOMEONE PLEASE WRITE A EQUATION OF THIS LINE!!
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Answer (<u>assuming it can be in slope-intercept form)</u>:

y = -x - 1

Step-by-step explanation:

When knowing the slope of a line and its y-intercept, you can write an equation to represent it in slope-intercept form, or y = mx + b format. Substitute the m and b for real values.

1) First, find the slope of the equation, or m. Pick any two points from the line and substitute their x and y values into the slope formula, m = \frac{y_2-y_1}{x_2-x_1}. I chose the points (0, -1) and (-1, 0):

m = \frac{(0)-(-1)}{(-1)-(0)} \\m = \frac{0+1}{-1-0} \\m = \frac{1}{-1} \\m = -1

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2) Now, find the y-intercept, or b. The y-intercept of a line is the point at which the line crosses the y-axis. By reading the graph, we can see that the line intersects the y-axis at the point (0,-1), therefore that must be the y-intercept.  

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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.

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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}

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Integrating both sides

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

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\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

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Answer:

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