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3 years ago

How do I solve this?

Mathematics

2 answers:

DerKrebs [107]3 years ago

4 0

( 25 x^4 y^6 )^ 1/2 =

√25 x^4 y^6 =

√5^2 × √x^4 × √y^6 =

5^2/2 × x^4/2 × y^6/2 =

5 × x^2 × y^3 =

5x^2 y^3

scZoUnD [109]3 years ago

3 0

Answer:

25x^2 y^3

Step-by-step explanation:

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What is the surface area of a box with the dimensions of length 4cm, width 3cm, and height 5cm? Only type the number - no units.

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Answer:

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Step-by-step explanation:

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3 years ago

Raymond bought 5 rolls of paper towels. He got 99 4/5 meters of paper towels in all. How many meters of paper towels were on eac

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Answer:

19.96

Step-by-step explanation:

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3/5 and 0.4. Which number is the smallest?

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0.4 is the answer, 3/5= 0.6

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3 years ago

A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

VMariaS [17]

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

-F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

F_n,b - m*g = 0

F_n,b = m*g

- The frictional force F_f is given by:

F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

- m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

6.6*cos(Q) = 4*u_s*sin(Q)

tan(Q) = 6.6 / 4*u_s

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tan(Q) = 6.6 / 4*0.23

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4 years ago

The Problem of the Points. Two players engage in a game. They are equally matched, and the chances of winning a point are the sa

Dominik [7]

Answer:

Player A receives $77.34375 of the sakes

Player B receives $22.65625 of the stakes

Step-by-step explanation:

Given that at the time the game was interrupted, we have;

The number of points player A has = 7 points

The number of points player B has = 5 points

The amount the first player to score 10 points wins = $100

The number of points remaining for player A to win the game = 3

The number of points remaining for player B to win the game = 5

Therefore, the number of trials remaining for a winner to emerge = 3 + 5 - 1 = 7 trials

We take the probability that Player A wins a point as success

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$P(S) = \sum\limits_{j=3}^7 \dbinom{7}{3} \cdot \dfrac{1}{2^7} = \dfrac{35}{128} + \dfrac{35}{128} + \dfrac{21}{128} + \dfrac{7}{128} + \dfrac{1}{128} = \dfrac{99}{128}$

Therefore, given that the likelihood of Player A winning = 99/128

The stakes should be divided such that Player A gets 99/128 share of the stakes while Player B gets 1 - 99/128 = 29/128 share of the stakes

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Player B gets (29/128) × $100 = $22.65625

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3 years ago