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mr_godi [17]
3 years ago
8

Consider angle θ in Quadrant ll, where cos θ= -12/13. what's the value of sin θ​

Mathematics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer:

sinθ = 5/13

Step-by-step explanation:

cosθ​ = adjacent/hypotenuse (or you could think of it as x/r)

cosθ = -12/13

adjacent = -12

hypotenuse = 13

opposite = ?

To get the opposite, we use the Pythagorean theorem

hypotenuse^{2} = adjacent^{2}  + opposite^2

hypotenuse^{2} - adjacent^{2}  = opposite^2

\sqrt(hypotenuse^{2} - adjacent^{2})  = opposite

Now sub in the numbers:

\sqrt{(13^2 - (-12)^2)} = opposite

opposite = 5

Now sinθ = opposite/hypotenuse

so sinθ = 5/13

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Meredith is saving up to buy a digital camera that costs $490. So far, she saved $175. She would like to buy the camera 4 weeks
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She would have to get 78.75 dollars every week to get the camera.

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Working together, Tyler and Evan painted a fence in 8 hours. If it takes Tyler 12 hours less than it takes Evan to paint the fen
Rudik [331]

Answer:

24hours

Step-by-step explanation:

let Evan be x, Tyler=x-12,equate;1/x+1/(x-12)=1/8,to get x=24 or 4.The correct value of x=24 because if you take x=4 means that Tyler=-8hours which is impossible

8 0
3 years ago
The size of angle aob is equal to 132 degrees and the size of angle cod is equal to 141 degrees. find the size of angle dob.
Varvara68 [4.7K]

angle AOB = 132 and is also the sum of angles AOD and DOB. Hence 
angle AOD + angle DOB = 132°    ---> 1 

angle COD = 141 and is also the sum of angles COB and BOD. Hence 
angle COB + angle DOB = 141°    ---> 2

Now we add the left sides together and the right sides of equations 1 and 2 together to form a new equation. 

angle AOD + angle DOB + angle COB + angle DOB = 132 + 141       ---> 3 

We should also note that: 
angle AOD + angle DOB + angle COB = 180° 

Therefore substituting angle AOD + angle DOB + angle COB in equation 3 by 180 and solving for angle DOB:
180 + angle DOB = 132 + 141 
angle DOB = 273 - 180 = 93° 

8 0
3 years ago
Find the difference (5a^2+4ab-3b^2)-(-5ab+4b^2+3a^2)
Vesna [10]

Answer:

\boxed{\bold{2a^2+9ab-7b^2}}

Step By Step Explanation:

Remove Parenthesis: (a) = a

\bold{5a^2+4ab-3b^2-\left(-5ab+4b^2+3a^2\right)}

Simplify \bold{-\left(-5ab+4b^2+3a^2\right)}

\bold{5ab-4b^2-3a^2}

Rewrite Equation:

\bold{5a^2+4ab-3b^2+5ab-4b^2-3a^2}

Simplify \bold{5a^2+4ab-3b^2+5ab-4b^2-3a^2}

\bold{2a^2+9ab-7b^2}

- Mordancy

8 0
3 years ago
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Torrance is shopping for a school party. His donation to the party is snack bags and juice boxes. Snack bags come in packages of
lara [203]

Answer:

Torrance must purchase 5 packages of snack bags and 6 packages of juice boxes.

Step-by-step explanation:

Given that:

Snack bags come in packages of 12.

Juice bags come in packages of 10.

To find:

Fewest number of packages of each product so that there are same number of snack bags and juice boxes.

Solution:

Number of snack bags when 1 package is bought = 12

Number of snack bags when 2 package is bought = 24

Number of snack bags when 3 package is bought = 36

Number of snack bags when 4 package is bought = 48

Number of snack bags when 5 package is bought = 60

Number of snack bags when 6 package is bought = 72

Number of juice bags when 1 package is bought = 10

Number of juice bags when 2 package is bought = 20

Number of juice bags when 3 package is bought = 30

Number of juice bags when 4 package is bought = 40

Number of juice bags when 5 package is bought = 50

Number of juice bags when 6 package is bought = 60

Number of juice bags when 7 package is bought = 70

We can see that when 5 packages of snack bags are bought and 6 packages of juice bags are bought, 60 bags of each are bought.

This can be found by finding the LCM as well.

LCM of 10 and 12 is 60.

It means we need to buy 60 bags of each item.

And we can easily find the number of packages for each.

Number of packages of snack bags to be bought = \frac{60}{12} =5

Number of packages of juice bags to be bought = \frac{60}{10} = 6

Torrance must purchase 5 packages of snack bags and 6 packages of juice boxes.

4 0
3 years ago
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