Answer:
8
Step-by-step explanation:
This problem uses the pythagorean theorem. 15 squared plus 8 squared equals 17 squared.
Answer:
For the triangle, its area is 32.1 inches.
For the trapezoid, its area is 170 feet.
Step-by-step explanation:
The formula for finding the area of a triangle is A = h * b/2, where h is for height and b is for base. There seems to be two measurements at the base of the triangle so what I did is I added together 3.5 and 7.2 and got 10.7. The slant measurement of the triangle isn't important in this formula so just pretend 5.5 in isn't there. Multiply 10.7 (base) to 6 (height) and you'll get 64.2. Now divide the product by 2 and you'll get 32.1 inches as the area of the triangle.
The formula for finding the area of a trapezoid is A = ((a + b)/2)h, where a is base A, b is base B, and h is the height. This formula looks a bit complicated but if you know a bit of PEMDAS, then you'll understand the formula. Again, there are two measurements at the base of the trapezoid but simply, I added them to get 24 and the slants don't make any use for this formula either. Add 10 (base A) and 24 (base B) to get 34. Now divide 34 by 2 and you'll get 17. Finally, multiply the product to 10 (height) and you'll get 170 feet as the area of the trapezoid.
If you don't know PEMDAS, just comment something related to "Can you explain what PEMDAS is?" or "What is PEMDAS?" and I will answer your question as quick as I possibly can. I hope this helps!
56) the answer is 3x+10
_____
5
If they want it simplified it will be 3x+2
57) x+40
____
5
If they want it simplified it would be x+8
I'm not sure about 58 sorry I hope this helps a little.
Answer:
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<u>Answer</u>
2nd diagram
<u>Explanation.</u>
When contraction a parallel line from a point say N, outside the line, the first thing is to draw a line from point N to the that line.
The point where this line from N intersect with line, name it say P. From this point you can use the properties of angles in a parallel lines to construct the parallel line.
The line NP can act like a transverse of the two parallel lines. The diagram 2 shows first step.
