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Flura [38]
3 years ago
15

**PLEASE HELP ASAP** Betsy is analyzing a quadratic function f(x) and a linear function g(x). Will they intersect?

Mathematics
1 answer:
Nina [5.8K]3 years ago
5 0
We analyze the chart and observe that the linear function is y=x, since this relation holds for all values in the table.  Drawing this line over the quadratic function shows that they intersect twice, at both the positive and negative x-coordinates.

This is by far the easiest way to solve this problem, but if you're interested in learning how to do it algebraically, read on!  To prove this more rigorously, we can find that the equation of the parabola is y=\frac13 x^2 - 2.  Substituting in y=x, we find that the intersection points occur where \frac13 x^2 - 2 = x, or \frac13 x^2 - x - 2 = 0, or x^2 - 3x - 6 = 0.

This equation doesn't factor nicely, so we use the quadratic formula to learn that x = \frac{3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-6)}}{2} = \frac{3 \pm \sqrt{33}}{2}.  Hence, the x-coordinates of the intersection points are \frac{3 + \sqrt{33}}{2}, which is positive, and \frac{3 - \sqrt{33}}{2}, which is negative.  This proves that there are intersection points on both ends of the axis.
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I need help with this!​
Inga [223]

Answer:

Slope = 3/5

Step-by-step explanation:

go over three and up five and the you have 3/5

6 0
3 years ago
The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate
KonstantinChe [14]

Answer:

It will take 47 hours

Step-by-step explanation:

We can have the exponential function as follows;

P = I( 1 + r)^n

if the initial value was x , the final will be 2x

The change rate is 1.5% , which is r, this is same as;

1.5/100 = 0.015

So what we want to calculate is the n attached

This will be;

2x = x ( 1 + 0.015)^n

2 = 1.015^n

ln 2 = n ln 1.015

n = ln 2/ln 1.015

n = 46.55

This is approximately 47 hours

7 0
2 years ago
Find the area of the rhombus if AE = 20 m and DE = 32 m
Fudgin [204]
We have to calculate the area of the rhombus if AE = 20 m and DE = 32 m.The diagonals DB = 2 * DE and AC = 2 * AE.
DB = 2 * 32 = 64 mAC = 2 * 20 = 40 mArea = d1 * d2 / 2 = DB * AC / 2 = 64 * 40 / 2 = 1,280 m²Answer: The area of the rhombus is B ) 1,280 m².
4 0
3 years ago
The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
2 years ago
In the diagram, polygon ABCD is reflected across to make polygon A'B'C'D'. Which statement must be true?
UNO [17]
True polygon is the abcd polygon
6 0
3 years ago
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