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3 years ago

A rectangle has a height of 6k3 and a width of 2k2 + 4k +5.

Mathematics

1 answer:

Trava [24]3 years ago

5 0

Answer: $12k^5+24k^4+30k^3$

Step-by-step explanation:

Given : The height of the rectangle = $6k^3$

The width of the rectangle = $2k^2+4k+5$

Formula : Area = height x width

Therefore , the area of triangle in terms of polynomial will be :

$6k^3\times( 2k^2+4k+5)\\\\= 6k^3(2k^2)+6k^3(4k)+6k^3(5)\ \ [\text{Using Distributive property}]\\\\=12k^{3+2}+24k^{3+1}+30k^3\ \ [\text{Using exponents rule}:\ a^n\times a^m=a^{n+m}]\\\\=12k^5+24k^4+30k^3$

Hence, the area of the entire rectangle = $12k^5+24k^4+30k^3$

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Help me with question 1 please

GrogVix [38]

Answer:

DB = 6

Step-by-step explanation:

See the attached figure to the question.

Since O is the center of the given circle, so AB and CD are diameters of the circle.

Hence, AO = CO = OD = OB = radius of the circle = 6

In Δ OAC, we have AO = OC. Hence, ∠ A = ∠C = 60° {Given that ∠OAC = 60°}

So, ∠ AOC will automatically become 60°. {As the sum of all the angles of a triangle is 180°}

So, Δ AOC is equilateral triangle.

Now, ∠ AOC = ∠ BOD = 60° {Since, they are vertically opposite angles}

Now, in Δ BOD, we have DO = OB. Hence, ∠ D = ∠ B = 60° {As the sum of all the angles of a triangle is 180°}

So, Δ BOD is also an equilateral triangle.

So, DB = BO = DO = 6. (Answer)

8 0

3 years ago

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 0.93. Assume indepen

noname [10]

Answer:

There is a 99.99998% probability that at least one valve opens.

Step-by-step explanation:

For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 5, p = 0.93$

Calculate P(at least one valve opens).

This is $P(X \geq 1)$

Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807$

Finally

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193$

There is a 99.99998% probability that at least one valve opens.

5 0

2 years ago

Read 2 more answers

There are 16 students who are participating in a chess tournament. The players will be randomly chosen. What is the probability

Ainat [17]

2/16 which is equal to 1/8 or 0.125

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2 years ago

Simplify square root of 2 over cube root of 2.

Fed [463]

Answer:

2 to the power of one sixth

Step-by-step explanation:

Assuming you don't already know this, any type of root can be expressed as an exponent. Generally speaking:

$\sqrt[n]{x} = {x}^{ \frac{1}{n} }$

So you can rewrite the given fraction as

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } }$

and then reduce as you normally would. That is, if the bases of the numerator and denominator are the same, then you can subtract the denominator's exponent from the numerator's exponent like so:

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } } = {2}^{ \frac{1}{2} - \frac{1}{3} }$