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yawa3891 [41]
2 years ago
15

Simplify the given expression. Write your answer with positive exponents. (x−4y3)−5

Mathematics
1 answer:
EastWind [94]2 years ago
4 0

Answer:   \frac{x^{20}}{y^{15}}

This is the same as writing (x^20)/(y^15)

===================================================

Work Shown:

\left(x^{-4}y^3\right)^{-5}\\\\\left(x^{-4}\right)^{-5}*\left(y^3\right)^{-5}\\\\x^{-4*(-5)}y^{3*(-5)}\\\\x^{20}y^{-15}\\\\\frac{x^{20}}{y^{15}}

In the second step, I used the rule that (a*b)^c = (a^c)*(b^c)

In step 3, I used the rule (a^b)^c = a^(b*c)

In the last step, I used the rule a^(-b) = 1/(a^b)

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-6p - 20 = -2p +4(1 - 3p)
kirza4 [7]

Answer:

p = 3

Hope this helps

7 0
3 years ago
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Which method would be more efiicient to solve the system y=1/2x and 2x+3y=28
Licemer1 [7]
Probably the subsitution method


y=1/2x
subsitute that fr y

2x+3(1/2x)=28
2x+3/2x=28
times 2 both sides
4x+3x=56
7x=56
divide by 7 both sides
x=8
sub back

y=1/2x
y=1/2(8)
y=4

(x,y)
(8,4)
6 0
3 years ago
Oh my gosh help me i
Ivanshal [37]

Sorry its hard to read and see

4 0
3 years ago
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Please help me for the brainliest answer
olga55 [171]

Answer:

3, 12, 27 and 300

Step-by-step explanation:

Plug n as 1, 2, 3 and 10.

3(1)² = 3

3(2)² = 12

3(3)² = 27

3(10)² = 300

8 0
3 years ago
EX 6.1 How many iterations will the following for loops execute? for (int i = 0; i < 20; i+ +) { } for (int i = 1; i <= 20
Romashka-Z-Leto [24]

Answer:

20, 20, 15, 20, 10, and 5 iterations respectively

Step-by-step explanation:

The loop "for (int i = 0; i < 20; i+ +)" starts at i=0 and increases the counter i by 1 at the end of each iteration. The counter must be less than 20, so the values of i in the loop are 0,1,2,3,...,19. These are 20 values, and thus the code iterated 20 times.

The loop "for (int i = 0; i < 20; i+ +)" starts at i=1 and increases the counter i by 1 at the end of each iteration. The counter can't exceed 20, so the values of i in the loop are 1,2,3,4,...,20. These are 20 values, and thus the code iterated 20 times.

Similarly, in "for (int i = 5; i < 20; i+ +)" i takes the 15 values are 5,6,7,...,19. Hence, the code iterated 15 times.

The loop "for (int i = 20; i > 0; i- -)" starts at i=20 and decreases the counter i by 1 at the end of each iteration, so the values of i in the loop are 20,19,18,...,1. These are 20 values, then we have 20 iterations.

The loop "for (int i = 1; i < 20; i = i + 2)" starts at i=1 and increases the counter i by 2 at the end of each iteration. The values of i in the loop are 1,3,5,7,...,19. These are 10 values, and thus the code iterated 10 times.

The loop "for (int i = 0; i < 20; i+ +)" starts at i=1 and multiplies the counter i by 2 at the end of each iteration, so the values of i in the loop are 1,2,4,8,16. Then we have 5 iterations.

6 0
3 years ago
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