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masya89 [10]
2 years ago
7

Simplify the expression 5^3 x 5^-5

Mathematics
2 answers:
Arlecino [84]2 years ago
5 0
D for sure I think so a lot 1.
kotykmax [81]2 years ago
4 0

Answer:

\frac{1}{5^2} --- 1 over 5 squared

Step-by-step explanation:

When multiplying terms with a common base, you just add the exponents:

x^a\times x^b=x^{a+ b}

That's true even when you don't have any exponents.

5\times5=5^1\times5^1=5^{1+1}=5^2=25

\rightarrow5^3\times5^{-5}\\\rightarrow5^{3-5}\\\rightarrow5^{-2}

A negative exponent isn't fully simplified, so there's another rule to use:

x^{-y}=\frac{1}{x^y}

That is '1 over x to the y' if it's too small to read.

\rightarrow5^{-2}=\frac{1}{5^2}

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Addition information need it like the picture
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Simplify<br> 1203<br> 52<br> 4 100<br> 551<br> 044<br> 5<br> 421<br> 50<br> 0<br> 127<br> 152
s2008m [1.1K]

Answer:

The answer is C

Step-by-step explanation:

I pick that answer because you have divide

8 0
3 years ago
Rex mixed x liters of fruit syrup with y liters of water to make a fruit punch. Three times the number of liters of water he mix
irga5000 [103]
3y = 2x + 6
x + y = 7

y = 7 - x
3(7 - x) = 2x + 6
21 - 3x = 2x + 6
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15 = 5x
15/5 = x
3 = x

x + y = 7
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y = 4

Solution is (3,4)

6 0
3 years ago
Read 2 more answers
Multiple choice/ please help!!!
Oxana [17]

Answer:

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4 0
3 years ago
PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0
3 years ago
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