We have
<span>3/5=(a+5)/25
3/5=(5/5)*(3/5)=15/25
therefore
15/25=</span>(a+5)/25<span>
15=a+5------------------------ > a=10
the answer is the option </span>a = 10
In proving that C is the midpoint of AB, we see truly that C has Symmetric property.
<h3>What is the proof about?</h3>
Note that:
AB = 12
AC = 6.
BC = AB - AC
= 12 - 6
=6
So, AC, BC= 6
Since C is in the middle, one can say that C is the midpoint of AB.
Note that the use of segment addition property shows: AC + CB = AB = 12
Since it has Symmetric property, AC = 6 and Subtraction property shows that CB = 6
Therefore, AC = CB and thus In proving that C is the midpoint of AB, we see truly that C has Symmetric property.
See full question below
Given: AB = 12 AC = 6 Prove: C is the midpoint of AB. A line has points A, C, B. Proof: We are given that AB = 12 and AC = 6. Applying the segment addition property, we get AC + CB = AB. Applying the substitution property, we get 6 + CB = 12. The subtraction property can be used to find CB = 6. The symmetric property shows that 6 = AC. Since CB = 6 and 6 = AC, AC = CB by the property. So, AC ≅ CB by the definition of congruent segments. Finally, C is the midpoint of AB because it divides AB into two congruent segments. Answer choices: Congruence Symmetric Reflexive Transitive
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Answer:
Step-by-step explanation:
To solve this problem, we need to multiple
and -2 by
and add the results together:




Adding the two together give the following:


When the equation is ax+b which is y =ax+b
Answer:
(a) square: L; triangle: 0.
(b) square: L·(-16+12√3)/11; triangle: L·(27-12√3)/11
Step-by-step explanation:
<u>Strategy</u>: First we will write each area in terms of its perimeter. Then we will find the total area in terms of the amount devoted to the square. Differentiating will give a way to find the minimum total area.
__
In terms of its perimeter p, the area of a square is ...
A_square = p^2/16
In terms of its perimeter p, the area of an equilateral triangle is ...
A_triangle = p^2/(12√3)
Then the total area of the two figures whose total perimeter is L with "x" devoted to the square is ...
A_total = x^2/16 + (L-x)^2/(12√3)
__
(a) We know when polygons are regular, the one with the most area for the least perimeter is the one with the most sides. Hence, the total area is maximized when all of the wire is devoted to the square.
__
(b) The derivative of A_total with respect to x is ...
dA/dx = x/8 -(L-x)/(6√3)
This will be zero when ...
x/8 = (L-x)/(6√3)
x(6√3) = 8L -8x
x(8 +6√3) = 8L
x = L·8/(6√3 +8) = 8L(6√3 -8)/(64-108)
x = L·(12√3 -16)/11
The total area is minimized when L·(12√3 -16)/11 is devoted to the square, and the balance is devoted to the triangle.