Answer:
<u>Given:</u>
- DC ║ AB
- CM = MB as M is midpoint of BC
i) <u>Since DN and BC are transversals, we have:</u>
- ∠DCM ≅ ∠NBM and
- ∠CDM ≅ ∠BNM as alternate interior angles
<u>As two angles and one side is congruent, the triangles are also congruent:</u>
- ΔDCM ≅ ΔNBM (according to AAC postulate)
So their areas are same.
ii)
<u>The quadrilateral has area of:</u>
- A(ADCB) = A(ADMB) + A(DCM)
<u>And the triangle has area of:</u>
- A(ADN) = A(ADMB) + A(NBM)
Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.
7=x this is for the first one but you have to 1
3
=
2
1
1
3
=
x
21
31=21x
2
1
⋅
1
3
=
2
1
⋅
2
1
21
⋅
1
3
=
21
⋅
x
21
21⋅31=21⋅21x
Multiply the numbers
Cancel multiplied terms that are in the denominator
7
=
2
1
⋅
2
1
7
=
21
⋅
x
21
7=21⋅21x
7
=
7
=
x
7=x
7
=
7
=
x
7=x
Answer:
x =40
Step-by-step explanation:
5= 1/5x-3
Add 3 to each side
5+3 = 1/5x -3+3
8 = 1/5x
Multiply by 5
8*5 = 1/5x *5
40 =x
<h3>
Answer: Choice B) 0.5</h3>
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Explanation:
For now, focus solely on the orange boxplot. The first quartile Q1 is the left edge of the box, which is at 3.5; while the value of Q3 is 7.5 (right edge of the box). The interquarile range (IQR) is ...
IQR = Q3 - Q1
IQR = 7.5 - 3.5
IQR = 4
This is basically the width of the box. Ignore the whiskers when it comes to the IQR.
Let A = 4 as we'll use it later.
---------------------------------
Find the IQR for the blue box plot
Q1 = 6 = left edge of the blue box
Q3 = 9.5 = right edge of the blue box
IQR = Q3 - Q1
IQR = 9.5 - 6
IQR = 3.5
Let B = 3.5 as we'll use it later
---------------------------------
Subtract the values of A and B to find the difference of the IQR values
A - B = 4 - 3.5 = 0.5
