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nlexa [21]
3 years ago
13

Describe the end behavior of a polynomial with a quadratic term with a negative coefficient and a linear term with a positive co

efficient.
Group of answer choices

a.falls to the left, rises to the right

b.rises to the left, falls to the right

c.falls to the left, falls to the right

d.rises to the left, rises to the right
Mathematics
1 answer:
iogann1982 [59]3 years ago
3 0

Answer: I think it’s A

Step-by-step explanation:

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The distance from city A to city B is approximately 2070 miles. A plane flying directly to city B passes over city A at noon. If
USPshnik [31]

Answer:

f(t) = -400t+2070

Step-by-step explanation:

According to the information of the problem the plane is traveling at a constant velocity therefore the function that models that gives the distance of the plane from city B at time t hours would be linear. Something like this.

f(t) = mt+b

The velocity is the slope of the line and if you take B as your reference you would be going backwards therefore we would say that   m = -400 . And as I mentioned before if you take B  as your reference your starting point would be 2070 miles away, therefore the function would be

f(t) = -400t+2070

3 0
3 years ago
What additional piece of information would prove △LMK≅△NMK by Side Angle Side?
laila [671]

Answer:

angle M

Step-by-step explanation:

if you are using side angle side the two sides may be

line LK and NK

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4 0
2 years ago
Read the word problem below.
MAVERICK [17]
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5 0
3 years ago
Read 2 more answers
X^2 + 5x = quadratic equation <br>=​
timama [110]

Hi there!

The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation. 

x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}  

x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}  

x = \frac{-5 \pm \sqrt{25 - 12} }{2}  

x = \frac{-5 \pm \sqrt{13} }{2}  

x = \frac{-5 \pm 3.61 }{2}  

x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}

x=-0.695 \ \textgreater \ \ \textgreater \  -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31

Therefore, the solutions to the quadratic equation  are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!

Your answer is 4.31

4 0
3 years ago
Please help :) !!!!!!!!!!!
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8. 70.5* 10. 30* 12. 5.1*
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