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algol13
3 years ago
9

Find the exact length of the third side. 82 18

Mathematics
1 answer:
masha68 [24]3 years ago
3 0

Answer:

80

Step-by-step explanation:

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Triangle MNO is an equilateral triangle with sides measuring 16 square root of 3 units.
Olenka [21]
M.N and O are 60 degree. And you have to know that 30 60 and 90 triangle. if 30 degrees opposite is measuring A , 60 is A square root 3 and the 90's opposite is being 2A. So 90 degree opposite is 16 square root 3 and 30 degree is being 8 square root 3 and 60 is
8 \sqrt{3}  \times  \sqrt{3}  = 24
the height is 24 .
6 0
3 years ago
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Which graph shows the solution to the system of linear inequalities?<br> y&lt; 2x-5<br> y&gt;-3x + 1
trapecia [35]

Answer:

See explanation

Step-by-step explanation:

Plot the solution sets to both inequalities.

1. For the inequality y First, plot the dotted line y=2x+5 (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

2\cdot 0-5=-5

so the origin does not belong to the needed part. Shade the part, which does not include origin.

2. For the inequality y>-3x+1. First, plot the dotted line y=-3x+1 (dotted because sign is without notion "or equal to"), then choose correct part by substitution coordinates of the origin.

-3\cdot 0+1=1>0

so the origin does not belong to the needed part. Shade the part, which does not include origin.

3. Find the common region of these two shaded parts - this is the solution to the system of two inequalities.

5 0
3 years ago
on field day, 1/10 of the students in Mrs. Brown's class competes in jumping events, 3/5 of the student competes in running even
hjlf
1/10 + 1/10 = 2/10
10 / 5 = 2
3/5 = 6/10
2/10 + 6/10 = 8/10
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6 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}

5 0
3 years ago
Plisss help on my assesment
Anon25 [30]

Answer:

The first one is equivalent [the 6x+48 = 2(3x+24)] and the second one is <u>NOT</u> equivalent [the 7x+21 ≠ 2(5x+3)]

Step-by-step explanation:

Just follow distributive property to solve these. You can ignore the first expression in both until you have to compare the answers.

1. 6x+48 and 2(3x+24)

2(3x+24) ---> 2(3x) + 2(24) ---> <u>6x + 48</u>

Bring in the first expression ~ <u>6x+48 and 6x+48 </u>

They are the same, so they are equivalent

2. 7x+21 and 2(5x+3)

2(5x+3) ---> 2(5x) + 2(3) ----> 10x + 6

Bring in the first expression ~ <u>7x+21 and 10x + 6</u>

They are NOT the same, so they are NOT equivalent

6 0
3 years ago
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