Answer:
CD = 14 cm; DE = 21 cm
Step-by-step explanation:
The perimeter is the sum of side lengths (in centimeters), so ...
CD + DE + CF + EF = 55
CD + DE + 8 + 12 = 55 . . . . . . . substittute for CF and EF
CD + DE = 35 . . . . . . . . . . . . . . subtract 20
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The segment DF is a diagonal of the rhombus, so bisects angle D. That angle bisector divides ΔCDE into segments that are proportional. That is, ...
CD/DE = CF/EF = 8/12 = 2/3
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So, we have two segments whose sum is 35 (cm) and whose ratio is 2 : 3. The total of "ratio units is 2+3=5, so each must stand for a length unit of 35/5 = 7 (cm). The sides are ...
CD = 2·7 cm = 14 cm
DE = 3·7 cm = 21 cm
<em>Check</em>
CD + DE = (14 +21) cm = 35 cm . . . . . matches requirements
We need two numbers that when multiplied will give you -40 and when added will give you -3:
(x+5) (x-8) this is your answer. And you want to know the process Here it goes. And attached is a chart I made you so you can visually see what this process is telling you:
1) identify a,b, and c in the trinomial ax^2 + bx+c (in your case it's 1x^2-3x-40= x^2-3x-40)
2) write down all factor pairs of c (numbers that multiplied produce -40)
3) identify which factor pair from the previous step sums up to b
4) Substitute factor pairs into two binomials
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
Answer:
The triangle has one solution. The remaining side c ≈ 4 and remaining angles B = 30°; C = 31°.
Option D is correct.
Step-by-step explanation:
if angle A is obtuse and if a > b then the triangle has one solution
We are given ∠ = 119° which is obtuse and side a= 7 and side b - 4 i.e 7>4 so, the triangle has one solution.
Finding remaining side c and ∠B and ∠C
Using Law of sines to find ∠B
a/sin A = b/sin B
7/sin 119° = 4/sin B
7 * sin B = 4 * sin 119
7*sin B = 4(0.874)
sin B = 3.496/7
B = sin^-1(0.4994)
B = 29.96 = 30°
We know that sum of angles of triangle = 180°
So, 180° = 119° + 30° +∠C
180° = 149° + ∠C
=> ∠C = 180° - 149°
∠C = 31°
Now finding c
b/sin B = c /sin C
4/Sin 30 = c/sin 31
4* sin 31 = c*sin 30
4*0.515 = c* 0.5
=> c = 4*0.515/0.5
c = 4.12 ≈ 4
So, Option D one solution; c ≈ 4; B = 30°; C = 31° is correct.
X= 12.5, this is because you need to add 25 to the other side of the equation to make 2x = 25 and then divide both sides by 2 to get x=12.5
