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Sveta_85 [38]
3 years ago
9

What is the solution for the system of equations? Use the substitution method to solve. y=5−3x5x−4y=−3 Enter your answer in the

boxes.
Mathematics
1 answer:
velikii [3]3 years ago
5 0

y=5-3x\\5x-4y=-3\\\\5x-4(5-3x)=-3\\5x-20+12x=-3\\17x-20=-3\\17x=-3+20\\17x=17\\x=17/17\\x=1\\\\y=5-3x\\y=5-3(1)\\y=5-3\\y=2\\\\5x-4y=-3\\5(1)-4(2)=-3\\5-8=-3\\-3=-3

So:

x=1

y=2

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By what percent will a fraction decrease if its numerator is decreased by 50% and its denominator is decreased by 25%?
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The answer is 66.6666%
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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
Pleaseeeee help is it -2 or 16 btw it’s geometry and it needs to be counterexample
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Answer:

false

Step-by-step explanation:

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A 17 foot piece of string is cut into two pieces so that the longer piece is 5 feet longer than twice the shorter piece. If the
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Answer:

Shorter piece= 4 ft, longer piece= 13 ft

Step-by-step explanation:

If shorter piece is x

Longer pc is 2x + 5

Together they make 17 feet

So x + 2x + 5 = 17

3 x + 5 = 17

3x = 17 - 5 = 12

3x = 12

X = 12/ 3 = 4

So shorter pc is 4 feet

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