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2 years ago

Why can you compare two fractions with the same denominator by only comparing the numerators

1 answer:

6 0

Think about it like this: You have two pizzas that are cut in 8 pieces. You ate 5 and your friend ate 7. Who ate more? Obviously your friend did since 7 is greater than 5. BUT, you can only do that if both pizzas were cut in 8 pieces. If your pizza was cut in 7 and his was cut in 12, you can't just look at the numerator and say your friend ate more. You have to find a common denominator. That's why when you have a common denominator you can compare fractions because the "things" are cut into equal pieces.

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

Please Complete the page.

Alexxandr [17]

7. -8/3
8. 4/35
9. -21/50
10. 35/64
11.-25/54

6 0

3 years ago

Suppose you mark n points on a circle, where n is a whole number greater than 1. The number of segments you can draw that connec

leonid [27]

The expression for the number of segments drawn out of the n number of points in the circle is,
n² / 2 - n / 2
Substituting directly to the expression the number of points, n, which is equal to 8,
8² / 2 - 8 / 2 = 28
Thus, there are 28 segments that can be drawn from the points.

5 0

2 years ago

The function, f(x) = –2x2 + x + 5, is in standard form. The quadratic equation is 0 = –2x2 + x + 5, where a = –2, b = 1, and c =

avanturin [10]

Given:

The function, f(x) = -2x^2 + x + 5

Quadratic equation: 0 = -2x^2 + x +5
where a = -2
b = 1
c = 5

The discriminate b^2 - 4ac = 41

To solve for the zeros of the quadratic function, use this formula:

x = ( -b +-√ (b^2 - 4ac) ) / 2a

x = ( 1 + √41 ) / 4 or 1.85
x = ( 1 - √41 ) / 4 or -1.35

Therefore, the zeros of the quadratic equation are 1.85 and -1.35.

4 0

3 years ago

Read 2 more answers

X squared divided 2=w-y make x the subject

Artemon [7]

Do you mean this?
x= sqrt{2(w-y)}

8 0

2 years ago