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2 years ago

In a triangle with vertices A,B and C, if C =64° and the lengths AC= 40mm and BC= 70mm find the length AB using cosine rule

Mathematics

1 answer:

boyakko [2]2 years ago

5 0

Answer:

63.6mm

Step-by-step explanation:

According to cosine rule;

AB² = BC²+AC²-2(BC)(AC)cos m<C

Substitute the given values

AB² = 70²+40²-2(70)(40)cos 64

AB² = 4900+1600-5600cos64

AB² = 6500-5600(0.4384)

AB² = 6500-2,454.87

AB² = 4,045.12

AB = √4,045.12

AB = 63.6mm

Hence the length of AB is 63.6mm

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krok68 [10]

Answer:

Part 1) The product of 3 and 2y-4x=8 is $6y-12x=24$

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Step-by-step explanation:

we have

$2y-4x=8$

step 1

Find out the product of 3 and 2y-4x=8

Multiply both sides by 3

$3(2y-4x)=3(8)$

Apply distributive property both sides

$6y-12x=24$

step 2

Find out the original equation in slope intercept form

The equation of the line in slope intercept form is

$y=mx+b$

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$2y-4x=8$

Solve for y

That means ----> Isolate the variable y

Adds 4x both sides

$2y=4x+8$

Divide by 2 both sides

$y=(4x+8)/2$

Simplify

$y=2x+4$

step 3

Find out the new equation in slope intercept form

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Solve for y

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Adds 12x both sides

$6y=12x+24$

Divide by 6 both sides

$y=(12x+24)/6$

Simplify

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