Answer: perimeter = 96mm, Area = 564mm².
Step-by-step explanation:
From the diagram, it is a trapezium, it has a rectangle joined together by a right angled triangle. Now to find the perimeter, we add all the dimensions together. To get the other side of the rectangle, which is also the perpendicular height of the figure, we take the Pythagoras. Let the perpendicular height = x
Therefore,
25² = ײ + 7²
625 = x² + 49
x² = 625 - 49
x² = 576
x. = √576
= 24.
Now the perimeter
= 20 + 25 + 7 + 20 + 24
= 96mm.
Area of the figure will be
= 1/2( a + b )h
Where a = 20, b = 20 + 7 = 27 and h = 24, now substitute for the values
( 20 + 27) x 24
---------------------
2
= 47 x 24/2
= 47 x 12
= 564mm²
Answer:
(-6127) + 0 = (-6127)
(+18) + 2 = (+20)
(+27) - 7 = (+20)
(+63) + (-13) = (+50)
(+21) - 4 = (+17)
Step-by-step explanation:
Answer:
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9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
