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cluponka [151]
3 years ago
8

Find E, the midpoint of RD, if R(4, –5) and D(2, 3).

Mathematics
1 answer:
dedylja [7]3 years ago
8 0

Answer:

E (3;-1)

Step-by-step explanation:

the 'x' coordinate of E: (4+2)/2=3;

the 'y' coordinate of E: (-5+3)/2= -1.

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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
Gnesinka [82]

\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

3 0
2 years ago
A window in the shape of a rectangle as shown below has a width of x+5 and a length of x^2- 3x+7 express the area of the rectang
Ksju [112]

Answer:

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Step-by-step explanation:

since a=lw, the area is (x^2-3x+7)(x+4)

1. distribute parentheses \left(x^2-3x+7\right)\left(x+4\right)=x^2x+x^2\cdot \:4+\left(-3x\right)x+\left(-3x\right)\cdot \:4+7x+7\cdot \:4

2. apply +(-a)=-a rule x^2x+4x^2-3xx-3\cdot \:4x+7x+7\cdot \:4

3. Simplify

    Steps to simplify:

x^2x=x^3

3xx=3x^2

3\cdot \:4x=12x

7\cdot \:4=28

x^3+4x^2-3x^2-12x+7x+28

4. Add like terms =x^3+4x^2-3x^2-5x+28

5. Add like terms =x^3+x^2-5x+28

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Whats 2*11 .... how was yall day?
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Answer:

22

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Answer:

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