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Crank
3 years ago
6

Help me with my work plz.

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

(√366 - 3)/24

Step-by-step explanation:

Given the following:

cos∝ = √3/8 and sinβ = √3/3

Sin(∝-β) = sin∝cosβ - cos∝sinβ

Get sin∝

Since cos∝ = √3/8

adj = √3

hyp = 8

opp = √8² - (√3)²

opp = √64 - 3

opp = √61

Recall that sin∝ = opp/hyp  

sin∝ = √61/8

Get cosβ

Since sinβ =  √3/3

opp = √3

hyp = 3

adj =√3² - (√3)²

adj = √9-3

adj = √6

Recall that cosβ = adj/hyp

cosβ = √6/3

Substitute the gotten values into the formula

Sin(∝-β) = sin∝cosβ - cos∝sinβ

Sin(∝-β) = ( √61/8)(√6/3)- (√3/8)(√3/3)

Sin(∝-β) = √366/24 - √9/24

Sin(∝-β) = (√366 - 3)/24

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Answer:

Equation of Hanger A:

x + y = z

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Equation for hanger B

x + y = p

p = 9 grams

Step-by-step explanation:

Note: This question is not complete, and lacks necessary data to answer this question. But for the sack of concept, we wil be solving this question using our own data.

Data given:

Triangle weight = 3 grams

Circle weight = 6 grams

There are two hangers A and B.

We have to find weight of the square in Hanger A and weight of the pentagon in Hanger B.

So, for your understanding I have drawn the two hangers A and B. In A we have triangle and circle on the LHS and the square on the right. On the other hand, we have pentagon on the right and triangle and circle.

Sketch is attached in the attachment.

Let's suppose: Hanger A is balanced, So,

Triangle weight = 3 grams = x

Circle weight = 6 grams = y

weight of the pentagon in Hanger B = p

weight of the square in Hanger A = z =?

So,

Equation of Hanger A:

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z = 9 grams  = weight of the square in Hanger A

Similarly for Hanger B.

Equation for hanger B

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