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Studentka2010 [4]
2 years ago
12

A,B and C are the vertices of a triangle,

Mathematics
2 answers:
Harlamova29_29 [7]2 years ago
3 0
SSS (side, side, side)
irinina [24]2 years ago
3 0

Answer:

SSS (side, side, side)

hope it helps:))!!!

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The line a bus arrives at the bus stop every 25 minutes, and the line b bus arrives every 15 minutes.Both are at the stop right
vova2212 [387]

Answer:

  after 75 minutes

Step-by-step explanation:

The least common multiple (LCM) of 15 and 25 is 75. It can be found a couple of ways:

1. List the factors of each number and find the product of the unique ones:

  15 = 3·5

  25 = 5²

The LCM is 3·5² = 75.

__

2. Find the greatest common divisor (GCD) and divide the product of the numbers by that value. From the above list of factors, we see that 5 is the GCD of 15 and 25. Then the LCM is ...

  15·25/5 = 75

__

Or, you can simply list multiples of each number and see what the smallest number is that is in both lists:

  15, 30, 45, 60, <em>75</em>, 90

  25, 50, <em>75</em>, 100

__

The two buses will appear together again after 75 minutes.

5 0
3 years ago
I need help with this as soon as possible
Arte-miy333 [17]
Yes the graph does show a proportional relationship because each x value has only one y value
5 0
3 years ago
Look at the screenshot, please. (no links)
elena-s [515]

Answer: 14

Step-by-step explanation: 18 x 14 = 252

3 0
3 years ago
Read 2 more answers
I need this please help mee ​
zubka84 [21]

Answer:

A, 120.

Step-by-step explanation:

x-15+75=180

x+60=180

x=120.

3 0
2 years ago
If sinθ = -1/2 and θ is in Quadrant III, then tanθ = _____.
Gnom [1K]

Answer:  \tan \theta=\dfrac{1}{\sqrt3}.

Step-by-step explanation:  Given that

\sin\theta=-\dfrac{1}{2} and \theta lies in Quadrant III.

We are to find the value of \tan \theta.

We will be using the following trigonometric identities:

(i)~sin^2\theta+\cos^2\theta=1,\\\\(ii)~\dfrac{\sin\theta}{\cos{\theta}}=\tan \theta.

We have

\tan\theta\\\\\\=\dfrac{\sin\theta}{\cos\theta}\\\\\\=\dfrac{\sin\theta}{\pm\sqrt{1-\sin^2\theta}}\\\\\\=\pm\dfrac{-\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\frac{\sqrt3}{2}}\\\\\\=\pm\dfrac{1}{\sqrt3}.

Since \theta lies in Quadrant III, so tangent will be positive.

Thus,

\tan \theta=\dfrac{1}{\sqrt3}.

8 0
3 years ago
Read 2 more answers
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