All sides are equal in an equilateral triangle and the perpendicular bisects the side too.
so BC will be bisected and the segments be 5 each.
you can use Pythagoras theorem to find the altitude length, hypotenuse will be one side of the triangle.

$h^2=100-25=75$
$h=\sqrt{75}=5\sqrt3$
Answer:
The number of times coin thrown was <u>500</u>.
Step-by-step explanation:
Given:
A coin lands on heads 200 times. the relative frequency of heads is 0.4.
Now, to find the times coin was thrown.
Let the number of times coin thrown was be 
Relative frequency = 0.4.
Number of lands on heads = 200.
So, to get the number of times coin was thrown we put formula:


<em>By cross multiplying we get:</em>

<em>Dividing both sides by 0.4 we get:</em>

Therefore, the number of times coin thrown was 500.
I thinks its 4.375 all u do is divide 35 by 8
This is easy if im right then it should be 1,000 because look at it this way they charge 55$ and they need to raise 55,000$ and what number has three zeros right off your 1,000 right so times that together and its 55,000
Answer:
The equation of the Parallel line to the given line is 5x+y-7=0
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given that the line y = - 5x+4 and point (1,2)
The equation of the Parallel line to the given line is ax+by+k=0
Given straight line 5x + y -4 =0
The equation of the Parallel line to the given line is 5x+y+k=0
This line passes through the point (1,2)
5x+y+k=0
5(1)+2+k=0
⇒ 7+k=0
k =-7
<u>Final answer:-</u>
The equation of the Parallel line to the given line is 5x+y-7=0