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SSSSS [86.1K]
3 years ago
9

What are the coordinates of the midpoint of the line segment with endpoints J(−6, 3) and K(4, −2) ?

Mathematics
1 answer:
34kurt3 years ago
3 0
Midpoint formula : (x1 + x2) / 2, (y1 + y2)/2
(-6,3)....x1 = -6 and y1 = 3
(4,-2)...x2 = 4 and y2 = -2
now sub
m = (-6 + 4) / 2, (3 - 2) / 2
m = (-2/2, 1/2)
m = (- 1, 0.5) <==
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Please someone help me. I'll give brainliest.
jekas [21]

Answer:

I would say the answer is C

3 0
3 years ago
Vanessa has scored 45, 32, and 37 on her three math quizzes. She will take one more quiz and she wants a quiz average of at leas
zhenek [66]

45 + 32 + 37 = 114

114÷3 = 38

4 total quizzes,


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for 3 quizzes she has 45 +32 +37 = 114 points


160-114 = 46, she needs at least a 46 on the 4th quiz


6 0
3 years ago
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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi
stealth61 [152]
Applying our power rule gets us our first derivative,

\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}

simplifying a little bit,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

looking for critical points,

\rm 0=2x^{-2/3}+4x^{1/3}

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

0=2x^{-2/3}\left(1+2x\right)

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

and take our second derivative.

\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}

Looking for inflection points,

\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}

Again, pulling out the smaller power of x, and fractional part,

\rm 0=-\frac43x^{-5/3}\left(1-x\right)

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
8 0
3 years ago
2 + 3(3x - 6) = 5(x - 3) + 15 I'm in Algebra 1 and I still can't seem to get this problem, I use Slader to find how to do this b
Wewaii [24]

Answer:

x = 4

Step-by-step explanation:

2 + 3(3x - 6) = 5(x - 3) + 15

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Subtract 5x from both sides:

4x - 16 = 0

Add 16 to both sides:

4x = 16

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7 0
3 years ago
What is the value of (Negative 14 Superscript 0 Baseline) Superscript negative 2? Negative StartFraction 1 Over 196 EndFraction
Nookie1986 [14]

Answer:

1

Step-by-step explanation:

Assuming, we want to find the value of ((-14)^0)^{-2}

Recall that: any non-zero number exponent zero is 1.

Using this property, we simplify our expression to (1)^{-2} since (-14)^0=1

Now using the property of exponents: a^{-n}=\frac{1}{a^n}

This implies that:1^{-1}=\frac{1}{1^2}=\frac{1}{1}=1

The correct answer is 1

5 0
3 years ago
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