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Sergio [31]
3 years ago
13

Triangles D E F and P Q R are shown. Given that StartFraction D F Over P R EndFraction = StartFraction F E Over R Q EndFraction

= three-halves, what additional information is needed to prove △DEF ~ △PQR using the SSS similarity theorem? DE ≅ PQ ∠D ≅ ∠P StartFraction D E Over E F EndFraction = three-halves StartFraction D E Over P Q EndFraction = three-halves
Mathematics
2 answers:
geniusboy [140]3 years ago
7 0

Answer:

D. DE/PQ= 3/2

Step-by-step explanation:

2020 edg

alukav5142 [94]3 years ago
3 0

Answer:

The correct option is;

DE/PQ = 3/2 which is StartFraction DE Over PQ Endfraction = three-halves

Step-by-step explanation:

The information given bout the triangles are;

DF/PR = FE/RQ = 3/2

Therefore since the given sides of triangle DEF are 3/2 times the sides of triangle PQR, the given sides of triangle DEF have been scaled 3/2 times to get the given sides of triangle PQR

Therefore, to prove that ΔDEF ~ ΔPQR, the ratio of the third side of triangle DEF, that is DE, to the third side of triangle PQR, which is PQ must be three-halves.

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The perimeter of a rectangle is 40cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm.
alexgriva [62]

Answer:

The dimensions of the original rectangle are Length =12cm, Width=8cm

Step-by-step explanation:

Let the length of the rectangle be x

Let the width of the rectangle be y

The perimeter of a rectangle is 40cm.

2(x+y)=40

Divide both sides by 2

x+y=20

If the length were doubled(2x) and the width halved(y/2), the perimeter would be increased by 16cm,i.e.(40+16)cm

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2(2x+\dfrac{y}{2})=56

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From the first equation, x=20-y.

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3 years ago
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Elina [12.6K]

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Answer:

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Arithmetic sequences have a common difference. Geometric sequences have a common ratio.

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The ratio is found by dividing a term by the one before it. If successive ratios are all the same, they are "common."

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<h3>2.</h3>

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<h3>3.</h3>

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<h3>4.</h3>

The common difference is -6: arithmetic.

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<h3>5.</h3>

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