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kykrilka [37]
3 years ago
7

Look at the addition problem above. Add the two fractions together, and write the answer below. *

Mathematics
1 answer:
MariettaO [177]3 years ago
8 0
9/10, this is just added on because Brainly won’t let me just send 9/10, please make my answer brainliest
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What is the approximate area of the shaded sector in the circle shown below?
Anton [14]

Answer:

The approximate area is A) 5.09cm²

Step-by-step explanation:

Area of a Circle = πr²

r = 1.8

Plug in our values

π(1.8cm²)

Evaluate the area of the entire circle

a = π(1.8cm²)

a = π(3.24cm)

a = 10.179cm²

Area of a sector, with the area of the circle of the sector being a = theta/360 * a

Evaluate the area of the sector

180/360*10.719cm²

0.5 * 10.179cm²

5.0895cm²

Round the value up

5.09cm²

3 0
2 years ago
Using addition formula solve tan 15​
salantis [7]

Answer:

2 - \sqrt{3}

Step-by-step explanation:

Using the addition formula for tangent

tan(A - B) = \frac{tanA-tanB}{1+tanAtanB} and the exact values

tan45° = 1 , tan60° = \sqrt{3} , then

tan15° = tan(60 - 45)°

tan(60 - 45)°

= \frac{tan60-tan45}{1+tan60tan45}

= \frac{\sqrt{3}-1 }{1+\sqrt{3} }

Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator.

The conjugate of 1 + \sqrt{3} is 1 - \sqrt{3}

= \frac{(\sqrt{3}-1)(1-\sqrt{3})  }{(1+\sqrt{3})(1-\sqrt{3})  } ← expand numerator/denominator using FOIL

= \frac{\sqrt{3}-3-1+\sqrt{3}  }{1-3}

= \frac{-4+2\sqrt{3} }{-2}

= \frac{-4}{-2} + \frac{2\sqrt{3} }{-2}

= 2 - \sqrt{3}

3 0
3 years ago
The six seventh grade classes held a fundraiser where they sold candy bars. The school's principal award a $1000 prize to be div
Gelneren [198K]

Answer:

8200/1000 = 8.2 so each time a class selss 8.2 bars, they receive 1 dollar.

If a class sold 16 bars, the class receives 2 dollars, if it sold 82 bars, it receives 10 dollars and so on.

6 0
2 years ago
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