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MrRa [10]
3 years ago
11

Help Please! Due Today In △ABC the points M and N belong to the sides AC and AB respectively. AM/MC=AN/NB=2/3. Prove that MN∥BC.

Find MN if BC=20.
Mathematics
2 answers:
sasho [114]3 years ago
5 0

Answer:

It does = 20 you're correct :)

Step-by-step explanation:

Ratling [72]3 years ago
4 0

Answer:

MN=8

Step-by-step explanation:

since AM/AC = AN/AB = 2/5,

then ∆ABC similar to ∆ANM.

since ∆ABC ~ ∆ANM,

then MN || BC.

by similarity

MN = BC•2/5 = 8

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lilavasa [31]

Those two functions just miss each other so there are no real solutions.  Every polynomial with complex coefficients has complex solutions.

2x^2 + 2 = 3x

2x^2 - 3x = -2

2( x^2 - \frac 3 2 x) = -2

x^2 - \frac 3 2 x = -1

x^2- \frac 3 2 x+ (\frac 3 4)^2= -1 + (\frac 3 4 )^2

(x - \frac 3 4)^2 = -\frac 7{16}

x - \frac 3 4 = \pm \sqrt{ -\frac 7{16}}

x = \frac 3 4 \pm \frac i 4 \sqrt{7}

x = \frac 1 4(3 \pm i \sqrt{7})


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Find the circumference of the circle to the nearest tenth. Use 3.14 for pi.
katrin2010 [14]

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Which ordered pair is a solution of the equation?<br> Y-4=7(x-6)
deff fn [24]

Answer:

(6,4)

Step-by-step explanation:

The equation is in 'point-slope' form.

y-y_1=m(x-x_1)\\\rule{150}{0.5}\\(x_1,y_1) \rightarrow \text{a point that's given}\\\\m \rightarrow \text{slope}

This means we can identify the point that was used in the equation.

y-\boxed{y_1} =m(x-\boxed{x_1}) | y-\boxed{4}=7(x-\boxed{6})\\\rule{210}{0.5}\\y_1 = 4\\\\x_1 =6\\\rule{210}{0.5}\\(x_1,y_1) \rightarrow \boxed{(6,4)}

(6,4) would be a solution to the equation given.

Hope this helps.

4 0
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