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adoni [48]
2 years ago
15

Simplify the radical, the square root of x^13

Mathematics
2 answers:
kotykmax [81]2 years ago
8 0
Rewrite it as a perfect square.

√x¹²x

Write it as a product of two radicals.

√x¹²√x

Simplify √x¹² (divide for 2)

√x¹² * 1/2

x⁶√x
myrzilka [38]2 years ago
5 0
√(x^13) is equal to:

(x^13)^(1/2)  which is equal to:

(x^12*x^1)^(1/2)  which is equal to

(x^12)^(1/2)*(x^1)^(1/2)

and the applicable rule:  (a^b)^c=a^(b*c) so

x^(12*1/2)*x^(1*1/2)

x^6*x^(1/2)

x^6√x
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A factory bottles 360 bottles of juice in 5 minutes. Find the factory's unit rate of bottles per minute.
castortr0y [4]

Find the unit rate by dividing 360 bottles by 5 minutes:

360 bottles

-------------------- = 72 bottles/minute

5 minutes

3 0
2 years ago
An object is dropped from a height of 1700 ft above the ground. The function h=-16t^2+1700 gives the object’s height h in feet d
Softa [21]
The answer is
<span>a) 1000=-16t^2+1700, implies t² = -700 /-16, and t= 6.61s
b) </span><span>970= -16t^2+1700, </span><span>implies t² = -730 /-16, and t=6.75s

c)
reasonable domain of h
h is polynomial function, so its domain is R, (all real number)
its range
the inverse of h is h^-1 = sqrt (1700- t / 16), and its domain is </span>
<span><span><span>1700- t / 16>=0, so t <1700,
the range of h is I= ]-infinity, 1700]</span> </span> </span>




8 0
2 years ago
Polygon CCC has an area of 404040 square units. Kennan drew a scaled version of Polygon CCC using a scale factor of \dfrac12 2 1
dsp73
The area of polygon D is given by:
 Ad = k ^ 2 * Ac
 Where,
 k: scale factor
 Ac: area of polygon C
 Substituting the values we have:
 Ad = (1/2) ^ 2 * (40)
 Ad = (1/4) * (40)
 Ad = 10 square units
 Answer:
 
the area of Polygon D is:
 
Ad = 10 square units
7 0
2 years ago
Read 2 more answers
PLEASE HELP ASAP: A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of z
dimaraw [331]
1) \frac{v(8)-v(0)}{8 - 0} = \frac{10-18}{8} = -1\frac{m}{s}.
2) v(5) = 5^2-9*5+18 = 25-45+18 = -2\frac{m}{s}.
3) The particle is moving right when the velocity function is positive: 0\ \textless \ t\ \textless \ 3 or 6\ \textless \ t\ \textless \ 8.
4) When 0\ \textless \ t\ \textless \ \frac{9}{2} the particle is slowing down because the acceleration is close to zero \Rightarrow the particle is speeding up when acceleration is increasing away from zero: \frac{9}{2}\ \textless \ t\ \textless \ 8.
5) (\frac{1}{8})* \int\limits^8_0 {t^2-9t+18dt}=\frac{1}{8}*(\frac{t^3}{3}-(\frac{9}{2}*t^2+18t)_{0}^{8}= \\=(\frac{1}{8})*(\frac{8^3}{3}-(\frac{9}{2})*8^2+18*8)=\frac{8^2}{3}-(\frac{9}{2})*8+18=3\frac{1}{3} \frac{m}{s}.
3 0
2 years ago
Which term of the AP. 8 , -4 , -16 , -28 ,......... is -880 ?​
denis-greek [22]

Answer:

75th term

Step-by-step explanation:

hope it is well understood

6 0
2 years ago
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