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Contact [7]
3 years ago
15

Is 7/8 greater than or less than 1/2

Mathematics
2 answers:
shutvik [7]3 years ago
6 0
Greater than. 1/2 is only 4/8
oksian1 [2.3K]3 years ago
4 0
7/8 is greater than 1/2
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if y equle 1 2 into 4. equle y into 1 ans 8

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Why would negative numbers not make sense for the values of the variables
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The equation of the line of best fit of a scatter plot is y = −7x − 2. What is the the y-intercept? (4 points)
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3 years ago
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Find sin(α) and cos(β), tan(α) and cot(β), and sec(α) and csc(β). The hypotenuse is 7 and side is 4.
Shalnov [3]

The triangle is missing, so i have attached it.

Answer:

1)sin(α) = 4/7

2)cos(β) = 4/7

3)tan(α) = 4/√33

4)cot(β) = 4/√33

5)sec(α) = 7/√33

6)csc(β) = 7/√33

Step-by-step explanation:

(1) sin(α)

From trigonometric ratios, we know that sine of an angle in a right angle triangle = opposite/hypotenuse.

Now, in this question, the opposite side to α is 4 and the hypotenuse is 7. Thus, sin(α) = 4/7

2) cos(β)

Cosine of an angle = adjacent side/hypotenuse.

In the question, the adjacent side to the angle β is 4 and the hypotenuse is 7. Thus, cos(β) = 4/7

3)tan(α)

tan of an angle = opposite/adjacent side. The opposite side to α is 4, but the adjacent side is unknown.

Using the pythagoras theorem,

Adjacent side = √(7² - 4²)

Adjacent side = √(49 - 16)

Adjacent side = √33.

Thus, tan(α) = 4/√33

4) cot(β)

cot of an angle is the reciprocal of tangent of same angle.

The adjacent side to β is 4 while the opposite is √33.

So, tan(β) = (√33)/4

cot(β) = 1/tan(β)

cot(β) = 1/[(√33)/4]

cot(β) = 4/√33

5)sec(α)

sec of an angle is equal to one divided by cosine of that same angle, so it equals hypotenuse divided by the adjacent. The hypotenuse is 7 and the adjacent side to α is √33.

Thus, sec(α) = 1/cosα = 7/√33.

6) csc(β)

Csc of an angle is equal to one divided by sine of same angle, so it equals hypotenuses divided by the opposite. The hypotenuse is 7 and the opposite side to β is √33.

Thus, csc(β) = 1/sin(β) = 7/√33

4 0
3 years ago
I need help with number 12&amp;13
IceJOKER [234]
Here's what I come up with hope it helps

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