Answer:
m < ANM = 36 degrees.
AM = 9.40 cm to the nearest hundredth.
Perimeter = 94.05 cm to the nearest hundredth.
Step-by-step explanation:
As we have a regular pentagon:
m < ANB = 360 / 5
= 72 degrees
So m < ANM = 1/2 * 72
= 36 degrees.
In the triangle ANM, AN = 16 so
sin 36 = Am / 16
AM = 16 sin36
= 9.4045 cm.
AB = 2 * AM = 18.809 cm
So as all sides are equal:
Perimeter = 5 * 18.809
= 94.05 cm.
If you plot the triangle on a graph, you'll see that the shape is a right triangle. Using the distance formula we can calculate the distance between point A and point B, which is the hypotenuse.
√<span><span><span>(<span>2− (−2)</span>)^</span>2 </span>+ <span><span>(<span>4−1</span>)^</span>2
</span></span>√<span><span><span>(<span>2+2</span>)^</span>2 </span>+ <span><span>(<span>4−1</span>)^</span>2
</span></span>√<span><span><span>(4)^</span>2 </span>+ <span><span>(3)^</span>2
</span></span>√<span><span>6+9
</span>√</span><span><span>25
</span>= 5
5 + 6 + 8 = 19. The perimeter of triangle ABC is 19 units. Hope this helps:)
~Ash</span>
Answer:
(25)^2*(9)^2*(4squreroot2)^2*(7squreroot2)^2
Step-by-step explanation:
5^4*3^3*2^5*343
=(25)^2*(9)^2*(4squreroot2)^2*(7squreroot2)^2
Answer:
The missing frequencies are x = 8 and y = 43.
Step-by-step explanation:
Median Value =70
Then the median Class =60-80
Let the missing frequencies be x and y.
Given: Total Frequncy = 200 , Median = 46
From the table
Here, n = 200
n/2 = 100
Lower Class Boundary of the median class, l=60
Frequency of the median class(f) =66
Cumulative Frequency before the median class, f=42+x
Class Width, h=10
200=149+x+y
200=149+8+y
y=200-(149+8)
y=43
Hence, the missing frequencies are x = 8 and y = 43.
What’s the picture of problem
