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hichkok12 [17]
3 years ago
11

3.

Mathematics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

A)

Step-by-step explanation:

length = x +x + 2 = 2x + 2

Width = 2+ x

Area of rectangle = length *width

(x +2)(2x + 2) = 110

Use FOIL method

x*2x + x*2 + 2*2x + 2*2 = 110

2x² + <u>2x + 4x</u> + <u>4 - 110</u> = 0   {Combine like terms}

2x² +6x - 106 = 0

a = 2 ; b = 6 ; c = -106

b² - 4ac = 6² - 4*2*(-106)

             = 36 + 848

             = 884

\sqrt{b^{2}-4ac}=\sqrt{884}= 29.73

x =\dfrac{-b+\sqrt{b^{2}-4ac}}{2a} \ ; \ x==\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}\\\\x = \dfrac{-6+29.73}{2*2} \ ; \ x =\dfrac{-6-29.73}{2*2} \ this \ is \  ignored \ as \ it \ is \  negative\\\\x=\dfrac{23.73}{4};

x = 5.93 m

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Domain of an ellipse is x ∈ [-a,a] and Range of an ellipse is y ∈ [-b,b].

So, given equation is an Ellipse where Domain is x ∈ [-5,5] and Range is y ∈ [-2,2].

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Answer:

12

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7 0
3 years ago
Anonymous 3 years ago
amm1812
Two angles are said to be complementary, if the sum of the measures of the to angles is equal to 90 degrees.

Thus, given that HFG is complementary to ACB, them mHFG + mACB = 90 degrees.

From the figure, given that the line from point F meats line CE at point P, then HFG = CFP.
But mCFE = 90 degrees and mCFE = mCFP + mPFE
Also PFE = DFH

Thus, mCFE = mCFP + mPFE = mHFG + mDFH = 90 degrees

Recall that mHFG + mACB = 90 degrees

Thus, mHFG + mACB = mHFG + mDFH

Therefore, mACB = mDFH.
5 0
4 years ago
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