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Marysya12 [62]
3 years ago
9

What substitution should be used to rewrite 16(x^3+1)^2-22(x^3+1)-3=0 as a quadratic equation?

Mathematics
2 answers:
Gnoma [55]3 years ago
7 0

Answer:B on edge

Step-by-step explanation:

lubasha [3.4K]3 years ago
3 0

16(x^3+1)^2-22(x^3+1)-3=0\\\\\text{substitute}\ t=x^3+1\\\\16t^2-22t-3=0

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Given: MNOK is a trapezoid, MN=OK, m∠M=60°, NK ⊥ MN , MK=16cm Find: The midsegment of MNOK
Sergeu [11.5K]

Answer:

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Step-by-step explanation:

1. Consider right triangle MNK. In this triangle angle N is right and m∠M=60°, then m∠K=30°. Thus, this triangle is special 30°-60°-90° right triangle with legs MN and NK and hypotenuse MK=16 cm. The leg MN is opposite to the angle with measure of 30°, then this leg is half of the hypotenuse, MN=8 cm.

2. Consider right triangle MNH, where NH is the height of trapezoid drawn from the point N. In this triangle m∠M=60°, angle H is right, then m∠N=30°. Similarly, the leg MH is half of the hypotenuse MN, MH=4 cm.

3. Trapezoid MNOK is isosceles, because MN=OK=8 cm. This means that NO=MK-2MH=16-8=8 cm.

4. The midsegment of the trapezoid is

\dfrac{MK+NO}{2}=\dfrac{16+8}{2}=12\ cm.

4 0
2 years ago
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Arada [10]

Answer:

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5 0
3 years ago
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What value must k take in order for the following expression to be greater than zero?
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Step-by-step explanation:

I think you meant 38 1/2

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but if you didn't then its

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