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3 years ago

PLEASE HELP I ASKED THIS QUESTION YESTERDAY AND NOBODY ANSWERED.

Mathematics

1 answer:

ehidna [41]3 years ago

4 0

Graph A is correct! :)

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HELP NOW!!!What is the value of x?

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Answer:

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4 years ago

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krek1111 [17]

Answer:

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3 years ago

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F(-4) for f (x) = -4x+9<br>f(-4)=?

Dovator [93]

F(x)= -4x+9

Multiply f*x

xF=9+4x

Solve for X: -4x+xF=9+4x-4x=0

Combine like terms: -4x+4x=0

-4x+xF=9+0= 9

-9+-4x +xF = 0.

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3 years ago

You have 2 types of bread, 3 types of meat, and 2 types of cheese. How many different sandwiches can you make if each sandwich h

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3 years ago

What is the area of the sector ?

worty [1.4K]

They forget to say "not to scale". I'm guessing this is trig because I don't see another way to do it.

Let's consider x=chord PQ first. By the Law of Cosines

$x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ$

We have an isosceles triangle formed by two radii of 9 cm and x=PQ. By the Law of Cosines again,

$x^2 = 9^2 + 9^2 - 2(9)(9) \cos B$

$x^2 = 162 - 162 \cos B$

$\cos B = \dfrac{162 - x^2}{162} = 1 - \dfrac{x^2}{162}$

$\cos B = 1 - \dfrac{34 - 30\cos 81^\circ}{162}$

$B = \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ }{162} \right)$

The area of the circle is the fraction given by the angle,

$A = \dfrac{B}{360^\circ} \pi r ^2$

$A \approx \dfrac{ \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ}{162} \right) }{360} (3.142)9^2$

$A\approx 24.7474332960707$

Answer: 24.7 sq cm

8 0

3 years ago