Answer:
Regular Octagon
Step-by-step explanation:
There are only three regular shapes that tessellate – the square, the equilateral triangle, and the regular hexagon. All other regular shapes, like the regular pentagon and regular octagon, do not tessellate on their own.
To solve this you would use the pythagorean theorem since the brace is making the frame look like two right triangles. The theorem states that for a triangle with a right angle, A^2+B^2=C^2. A and B are the sides of the frame and C is the brace which is like the hypotenuse of the triangle. It doesn't matter which side is A or B so you can put 6 or 8 in place of either in the equation. 6^2+8^2=C^2. If you simplify this it equals 36+64=C^2, which then simplifies to 100=C^2. Then you take the square root of both sides (what number multiplied by itself = the number you are trying to get, in this case, 100). So then you get C=10 because 10x10=100. So the length of the diagonal brace is 10ft.
Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
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now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P
Solve for ppp. 16-3p=\dfrac23p+516−3p= 3 2 p+516, minus, 3, p, equals, start fraction, 2, divided by, 3, end fraction, p, plus
Ksivusya [100]
Given:
The given equation is:
To find:
The value of p.
Solution:
We have,
Multiply both sides by 3.
Isolating the variable terms, we get
Divide both sides by 11, we get
Therefore, the required solution is 
.
Answer:
2x-3 = 2x-5
because both sides have the same x value ( 2x) this problem has no solution
Step-by-step explanation:
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