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ludmilkaskok [199]
3 years ago
7

Find the angles marked with the letters.​

Mathematics
1 answer:
OLga [1]3 years ago
5 0

Answer:

bigger b is 55°, the other b is 35°, c is 29°

Step-by-step explanation:

angles opposite each other on a point are the same so the bigger b is 55

55+90=145

180-145=35 so b is 35

alternate angles are the same so 35+116=151

180-151=29 so c=29

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A wildlife shelter is home to birds, mammals, and reptiles. If cat chow is sold in 30 lb bags, what is the least number of bags
kotykmax [81]

Answer:

26 bags

Step-by-step explanation:

A bag of cat chows = 30 lbs

Amount of cat chows needed in a week = 15 lbs = ½ bag of cat chow

There are at least 52 weeks in a year.

The least number of cat chows bags needed at the shelter in 1 year = 52 weeks × ½ bag of chow

= 52 × ½

= 52/2

= 26 bags of cat chow

At least, 26 bags would be needed in 1 year (52 weeks) at this wild life shelter if 15 lbs is consumed weekly.

4 0
3 years ago
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I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

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3 years ago
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Answer:

Step-by-step explanation:

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QUESTION 9 of 10: Suppose cash income for the first three months is $200, 5850, and $1,060. Total expenses for these same three
Alenkasestr [34]

Answer:

c) $4,000,-$2,850, $5,170

Step-by-step explanation:

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10 spelling and 16 vocab because 10 * 2 is 20 and 16 * 5 is 80 and 80+20 is 100
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