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3 years ago

5

What is the greatest perfect square in 650

1 answer:

Ne4ueva [31]3 years ago

3 0

25*25=625. The answer is 25.

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MATH Intervention Math - Unit 2: Expressions, Sequences, and Logic

scZoUnD [109]

Huh this gives out no information to answer

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3 years ago

What are these two equations in slope intercept form?

Shtirlitz [24]

1. 4x - y = 8
4x = y + 8
Y = 4x - 8

2. 2x + 3y = 4
3y = -2x + 4
Y = -2/3x + 4/3

4 0

3 years ago

Read 2 more answers

What is the relationship between the numerator and the denominator in a fraction?

VLD [36.1K]

Don't really but I know that Google would help

4 0

3 years ago

In a particular game of chance, a wheel consists of 42 slots numbered 00, 0, 1, 2,...,40. To play the game, a metal ball is

Nikolay [14]

Answer:

The sample space for one spin of this game is: S = {00, 0, 1, 2, 3,…, 40}

Step-by-step explanation:

Consider the provided information.

In probability theory, the set of all possible outcomes or outcomes of that experiment is the sample space of an experiment or random trial. Using set notation, a sample space is usually denoted and the possible ordered outcomes are identified as elements in the set.

Here the possible number of elements in the set are 00, 0, 1, 2,...,40

The sample space is anything the ball can land on.

Thus, the sample space for one spin of this game is: S = {00, 0, 1, 2, 3,…, 40}

6 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago