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Molodets [167]
2 years ago
12

Lliana is painting a picture. She has green, red, yellow, purple, orange, and blue paint. She wants her painting to have four

Mathematics
2 answers:
Sedaia [141]2 years ago
6 0

Answer:  10

Step-by-step explanation:

Given : lliana is painting a picture. She has green, red, yellow, purple, orange, and blue paint.

Total number of colors = 6

If order does not matter, then we use combinations.

The number of combinations of n things taken r at a time :-

^nC_r=\dfrac{n!}{r!(n-r)!}

If green is already selected , then the remaining number of colors to choose  = 3 out of 5  ( red, yellow, purple, orange, and blue paint.)

^5C_3=\dfrac{5!}{3!(5-3)!}\\\\=\dfrac{5\times4\times3!}{3!2!}=10

The number of ways can she pick four colors if green must be one of them = 10.

atroni [7]2 years ago
3 0

Answer: i believe the answer is 4

Step-by-step explanation:

she can have

green red yellow purple

green yellow purple orange

green red purple orange

green red yellow orange

i couldn't see anymore combos but i could be wrong

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Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

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Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

\displaystyle \large{ \pi -  \frac{ \pi}{3}  =  \frac{3 \pi}{3}  -  \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{2 \pi}{3} }

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi  +   \frac{ \pi}{3}  =  \frac{3 \pi}{3}   +   \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{4 \pi}{3} }</u>

Both values are apart of the interval. Hence,

\displaystyle \large \boxed{ \theta =  \frac{2 \pi}{3} , \frac{4 \pi}{3} }

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Isolate sin(4 theta).

\displaystyle \large{sin 4 \theta =  -  \frac{1}{ \sqrt{2} } }

Rationalize the denominator.

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The problem here is 4 beside theta. What we are going to do is to expand the interval.

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Multiply whole by 4.

\displaystyle \large{0 \times 4 \leqslant  \theta \times 4 < 2 \pi \times 4} \\  \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

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<u>\displaystyle \large{ \pi +  \frac{ \pi}{4}  =  \frac{ 4 \pi}{4}  +  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{  \frac{5 \pi}{4} }</u>

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\displaystyle \large{ \frac{5 \pi}{4}  + 2 \pi =  \frac{13 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 4\pi =  \frac{21 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 6\pi =  \frac{29 \pi}{4} }

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\displaystyle \large{ \frac{ 7 \pi}{4}  + 2 \pi =  \frac{15 \pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 4 \pi =  \frac{23\pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 6 \pi =  \frac{31 \pi}{4} }

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