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Nina [5.8K]
3 years ago
7

................................................

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

Answer:

it looks like if you multiply the small area by 4 you get the bigger area sorry thats all i know

Step-by-step explanation:

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Identify the pattern then write the next three terms in each sequence 192,96,48,24,...
Karo-lina-s [1.5K]
1st. Find the pattern
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4 0
2 years ago
How do you solve f(x)= 4x(2)-4x-3?
slavikrds [6]
If by 4x(2) you mean 4x to the power 2 or 4x^2...it can be solved in the following way
f(x)=4x^2-4x-3
=4x^2+2x-6x-3
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8 0
3 years ago
Can someone help me with this? Please this would mean so much.
Paha777 [63]

Answers:

x = angle MLN = 35 degrees

angle FLJ = 55 degrees

=================================================

Explanation:

This problem is fairly tricky if you're not sure what to look for. A slight clue is that they've marked a red point that strangely doesn't have a label on it. It's a fairly small point but it's definitely there if you look closely. While this point's location isn't exactly what we want, we're fairly close. Start at point L and draw a ray through the center point P. Ray LP will intersect the circle at point A, as shown in the diagram below.

From here, draw segments FA and MA. Now notice that inscribed angle LMF = 55 and inscribed angle FAL both subtend the same arc. The term "subtend" basically means "cut off". This arc that the inscribed angles subtend is minor arc FL. A minor arc is where you travel the shorter path around the circle, which indicates its measure is less than 180 degrees.

Since inscribed angles LMF and FAL subtend the same minor arc, this makes the inscribed angles to be congruent.

In short: angle FAL is 55 degrees

----------------------

Segment LA goes through the center P. Through Thales theorem, we know that inscribed angle LFA is 90 degrees. Consequently, we can determine that inscribed angle FLA is 90-55 = 35 degrees.

Segment JL is tangent to the circle, meaning that angle ALJ is 90 degrees. So angle FLJ is 90-35 = 55 degrees.

It's not a coincidence that angle FLJ, angle FAL, and angle LMF are the same measure.

----------------------

We found that angle FAL was 55 degrees. Applying Thales theorem again shows that angle MAF is 90 degrees. Therefore, angle LAM is 90-55 = 35 degrees.

Focus now on triangle LMA. This is also a right triangle (Thales Theorem). The upper acute angle we found was angle LAM = 35, so the lower acute angle is ALM = 55.

Then we can find angle MLN = (angle ALN) - (angle ALM) = 90 - 55 = 35

In short, angle MLN = 35 degrees

Similar to the previous section, it is not a coincidence that angles LFM, LAM and MLN are the same measure.

------------------------

As an alternative, since we know angle FLJ = 55, and angle FLM is 90 degrees, this means...

(angle FLJ)+(angle FLM) + (angle MLN) = 180

55 + 90 + angle MLN = 180

145 + angle MLN = 180

angle MLN = 180 - 145

angle MLN = 35 degrees

7 0
2 years ago
Please help with this simple trig problem- angle of two sectors of a circle
igomit [66]

Answer:

  K = (1/2)r^2(sin(θ) +θ)

Step-by-step explanation:

The area of the triangle to the left is ...

  A1 = (1/2)r^2·sin(180°-θ) = (1/2)r^2·sin(θ)

The area of the sector to the right is ...

  A2 = (1/2)r^2θ

so the total area of the blue shaded region is ...

  K = A1 + A2 = (1/2)r^2·sin(θ) + (1/2)r^2·θ

  K = (1/2)r^2(sin(θ) +θ)

4 0
3 years ago
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