Step-by-step explanation:
Given:
Complementary angle = 68'
Angle measure = 90' - 68' = 22'
Supplementary angle = 180' - 22' = 158'
Angle classification = Acute angle
Given:
Supplementary angle = 161'
Angle measure = 180' - 161' = 19'
Complementary angle = 90' - 19' = 71'
Angle classification = Acute angle
Given:
Angle measure = 136'
Angle classification = Obtuse angle
Complementary angle = 136' - 90' = 46'
Supplementary angle = 180' - 136' = 44'
2×5^{2} if you don't understand its 2 times 5 to the second power.
The points you are looking for are the midpoints of segments JL and JK.
J(-2, -1), K(4, -5), L(0, -5)
The midpoint of segment JL is
(-2 + 0)/2, (-1 + (-5))/2) = (-2/2, -6/2) = (-1, -3)
The midpoint of segment JK is
(-2 + 4)/2, (-1 + (-5))/2) = (2/2, -6/2) = (1, -3)
Answer: The coordinates are (-1, -3), (1, -3)
Answer:
Below
Step-by-step explanation:
All figures are squares. The area of a square is the side times itself
Let A be the area of the big square and A' the area of the small one in all the 5 exercices
51)
● (a) = A - A'
A = c^2 and A' = d^2
● (a) = c^2 - d^2
We can express this expression as a product.
● (b) = (c-d) (c+d)
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52)
● (a) = A-A'
A = (2x)^2 = 4x^2 and A'= y^2
● (a) = 4x^2 - y^2
● (b) = (2x-y) ( 2x+y)
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53)
● (a) = A-A'
A = x^2 and A' = y^2
● (a) = x^2-y^2
● (b) = (x+y) (x-y)
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54)
● (a) = A-A'
A = (5a)^2 = 25a^2 and A' =(2b)^2= 4b^2
● (a) = 25a^2 - 4b^2
● (a) = (5a-2b) (5a+2b)
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55)
● (a) = A - 4A'
A = (3x)^2 = 9x^2 and A'= (2y)^2 = 4y^2
● (a) = 9x^2 - 4 × 4y^2
● (a) = 9x^2 - 16y^2
● (a) = (3x - 4y) (3x + 4y)
Answer:
-24degC
Step-by-step explanation:
-6degC dropping by 2degC leads to -8degC (2degC colder, higher value negative numbers are colder as negative numbers increase in the opposite direction of positives)
so next -8degC rising by 3degC which is 3degC hotter(less negative) will give -5degC
the finap drop by 9degC makes the final temperature -5 -19 = -24degC(similar reasoning)
if you want a more straightforward method to doing these sorts of questions, just take temperature rise : add the value it rose by and temperature fall/drop: subtract the value it dropped/falled by.
