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saul85 [17]
3 years ago
12

Which answer would be correct

Mathematics
1 answer:
tatuchka [14]3 years ago
5 0
The answer I believe is b

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Please answer I need it fast thx ¡,,
Alla [95]

Answer:

whats your question? I can't see what you need

7 0
4 years ago
Help please, need to show work.
stiks02 [169]

9514 1404 393

Answer:

  3.  m∠A < 80°

Step-by-step explanation:

The exterior angle NGS, marked as 80°, has a measure that is the sum of the measures of the remote interior angles: GAN, GNA.

If we assume that angle measures cannot be zero, then we must have ...

  m∠A +m∠N = 80°

  m∠A = 80° -m∠N

Since m∠N > 0, the measure of angle A must be less than 80°.

This conclusion matches choice 3.

8 0
3 years ago
What number would be 10 times more than 3162​
marissa [1.9K]

Simply multiply 3162 times 10 to find that 31620 is the answer.

4 0
3 years ago
Read 2 more answers
Mark practices his guitar 3/4 hours each day. How many hours does he practice in seven days?
Mashutka [201]
3/4 *7
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6 0
4 years ago
Lim x-1 x³-2x²+3x-2/(2x^4-3x+1)
UkoKoshka [18]

Since the limit becomes the undetermined form

\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}

it means that both polynomials have a root at x=1. So, we can fact both numerator and denominator:

x^3-2x^2+3x-2 = (x-1)(x^2-x+2)

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

So, the fraction becomes

\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}

Now, as x approaches 1, you have no problems anymore:

\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}

4 0
3 years ago
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