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Alex73 [517]
4 years ago
8

Water flows over a dam at a rate of 500 gallons per minute. Write a function rule relating the amount of water (A) that flows ov

er the dam to the number of minutes (M) that have passed. What are the initial value and rate of change?
Mathematics
1 answer:
Bond [772]4 years ago
8 0
The function would be A(m) = 500m.  The initial amount of water flowing over the dam would be 0 and the rate of change would be 500.
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What is between 68 2/3 and 72 1/2 (options) 68 3/10, 747/11, 68 1/2, 550/8
enyata [817]
Why not write all of the numbers in decimal fraction form, to make them easier to compare?

68 2/3 = 68.666...
72 1/2 = 72.500
68 3/10 = 68.300
74 7/11 = 74.636...
68 1/2  = 68.500
550/8 = 68.750

Now ask yourself:  Which of the last four numbers is between 68.666.. and 72.500?
3 0
3 years ago
Read 2 more answers
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

6 0
3 years ago
Find a constant c for which p(z ≥ c.= 0.1587. round the answer to two decimal places.
Vikki [24]
What you see in the z score table is P(z< a constant), for P( z≥ a number), subtract the probability from 1:
1-0.1587=0.8413
on the z score table, you will see that p(z<1.00)=0.8413
so c=1.00
8 0
3 years ago
What is the slope of the line shown in the graph? <br> A.-3/2<br> B.-1/2<br> C.3/2
Varvara68 [4.7K]
I think answer is A
Because slope=rise/run
Rise is 3
Run is 2
But it will be negative because of the direction of slope
6 0
3 years ago
Will give brainlist to the best answer if possible!! 10 points please help
Black_prince [1.1K]

Answer:

B

Step-by-step explanation:

y= -x+18 and y= 3/4x+4

use slope intercept form: y=mx+b

m = slope

b = y-intercept

4 0
2 years ago
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