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MAVERICK [17]
3 years ago
5

△ABC∼△DEF , △ABC has a height of 14 centimeters, and △DEF has a height of 6 centimeters. What is the ratio of the area of △ABC t

o the area of △DEF ?

Mathematics
2 answers:
Kamila [148]3 years ago
8 0
Since triangle ABC is similar to triangle DEF then the ratio of the corresponding sides is constant.
The ratio of the corresponding lengths is referred to as the linear scale factor.
Considering the heights of the two triangles;
L.S.F = 14/6
          = 7/3
The ratio in area (A.S.F) is given by (L.S.F)²
Therefore, A.S.F = (7/3)² = 49/9
Thus te ratio of the area of triangle ABC to DEF is 49:9
Minchanka [31]3 years ago
5 0

Answer:

49:9

Step-by-step explanation:

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Answer: 1

Step-by-step explanation:

3x + 8x-8= 3

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3 0
3 years ago
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MakcuM [25]

Answer:

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Step-by-step explanation:

8 0
2 years ago
This is 15 points plz help ( no links ) EXPLAIN YOUR ANSWER plz
goldfiish [28.3K]
A.
EXPLANATION:
The mode is the peak of the data, in this data set, 5/8
The median is the center of data, in the data set, 1/2
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3 years ago
I need help with my math work
Gre4nikov [31]

Answer:

Go to tutoring, ask family, ask teacher, ask friend, or ask BRAINLY

Step-by-step explanation:

3 0
3 years ago
Solve the equation using the quadratic formula : x^2-10x+25=18
Sever21 [200]
x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}

x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}


Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}


8 0
3 years ago
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