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2 years ago

Set X = {x|x is a whole number less than or equal to 10} and set Y = {/15, 10, 15, 20}

Mathematics

1 answer:

SIZIF [17.4K]2 years ago

7 0

Answer:

Step-by-step explanation:

The first part requires the intersection of the two sets, which means only the elements that can be found in both sets.

While the second part requires all the elements in both the sets.

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Transversal relations

Shalnov [3]

Answer:

The answer is 75

Step-by-step explanation:

By way of parallelism

$x + 15 = 90 \\ x = 75$

7 0

2 years ago

Read 2 more answers

. The product of 4.7 and 6.5 equals 30.55. What is the product of 4.7 and 4.7 and

7nadin3 [17]

The answer is 3.5 & 355 because the decimal shifts to the left or right, you move the decimal in the answer too.

4 0

3 years ago

What is the answer to this equation?<br> 1/2 (4i+8)=-2

Flauer [41]

Answer:

If your looking for I the answer is -3

Step-by-step explanation:

Simplify both sides of the equation

1/2(4i+8)=-2 - Distribute

(1/2) (4i) + (1/2) (8) = -2

The subtract 4 from both sides

Then divide both sides by 2

5 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

2 years ago

Given a line with equation 3x +7y = 21. write an equation in point-slope form of the line parallel to the given line through the

guapka [62]

Answer:

The equation of the line is y = -3/7x - 23/7

Step-by-step explanation:

To start we need to find the slope of the original line. We can do this by solving the equation for y.

3x + 7y = 21

7y = -3x + 21

y = -3/7x + 3

Now we know the slope to be -3/7. Since parallel lines have same slopes, we know the new line also will have a -3/7 slope. We can now use that along with the given point in point-slope form to find the equation.

y - y1 = m(x - x1)

y + 5 = -3/7(x - 4)

y + 5 = -3/7x + 12/7

y = -3/7x - 23/7

8 0

2 years ago