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3 years ago

Find five consecutive integers whose sum is 195

2 answers:

7 0

$n;\ n+1;\ n+2;\ n+3;\ n+4-five\ consecutive\ integers\\\\(n)+(n+1)+(n+2)+(n+3)+(n+4)=195\\n+n+1+n+2+n+3+n+4=195\\5n+10=195\ \ \ \ \ |subtract\ 10\ from\ both\ sides\\5n=185\ \ \ \ \ \ |divide\ both\ sides\ by\ 5\\n=37\\\\Answer:\boxed{37;\ 38;\ 39;\ 40;\ 41}$

dimaraw [331]3 years ago

6 0

To do this, come up with three numbers. These are n, n+1, n+2, n+3, and n+4.

To solve, you do this:

$n+n+1+n+2+n+3+n+4=195 \\ 5n+10=195 \\ 5n+(10-10)=(195-10) \\ 5n=185 \\ \frac{5n}{5} = \frac{185}{5} \\ n=37$

Then, substitute 37 into the numbers:

n=37
n+1=37+1=38
n+2=37+2=39
n+3=37+3=40
n+4=37+4=41

The five consecutive integers are 37, 38, 39, 40, and 41.

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If you had a billion dollars and you spend one dollar every second how many years would it take you to spend it all

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There are 86,400 seconds in a day. 24 (hr) x 3,600 (sec)

Now we need to find how many seconds are in a year.

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365 x 86,400 = 31,536,000

Whew! Now that we have that out the way, we can now divide $1 billion by 31,536,000 seconds.

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3 years ago

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Let a, b, c, and d be real numbers with a, c 6= 0. Prove that the lines y = ax+b and y = cx + d have the same x-intercept if and

Monica [59]

Step-by-step explanation:

We have got the lines :

$y=ax+b\\y=cx+d$

Both lines intercept the x-axis in the point :

$I = (i_{1} ,i_{2})$

In all point from x-axis the y-component is equal to 0.

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We replace the I point in the lines equations:

$0=a(i_{1})+b \\0=c(i_{1})+d$

From the first equation :

$0=a(i_{1})+b \\-b=a(i_{1})\\i_{1}=\frac{-b}{a}$

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$0=c(i_{1})+d\\ -d=c(i_{1})\\i_{1}=\frac{-d}{c}$

Then $i_{1}=i_{1}$

Finally :

$\frac{-b}{a}=\frac{-d}{c} \\\frac{b}{a}=\frac{d}{c} \\ad=bc$

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) A watershed experiences a rainfall of 8 inches. What is the runoff volume when the curve number is 80

finlep [7]

Answer:

5.625 inches

Step-by-step explanation:

Given that:

Total Rainfall in inches (P) = 8 inches

The runoff volume (in inches) Q = ???

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Recall that: The runoff volume can be calculated by using the formula:

$Q = \dfrac{(P-0.2S)^2}{(P+0.8S)}$ for P > 0.2S

Q = 0 for P < 0.2S

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S = 2.5 inches

Since the rainfall (P) is greater than 0.25

Then:

$Q = \dfrac{(P-0.2S)^2}{(P+0.8S)}$

$Q = \dfrac{(8-0.2(2.5))^2}{(8+0.8(2.5))}$

$Q = \dfrac{(8-0.5)^2}{(8+2)}$

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$Q = \dfrac{(56.25)}{(10)}$

Q = 5.625 inches

Thus, the runoff volume = 5.625 inches

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