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wariber [46]
3 years ago
14

If a car is traveling east when it passes the camera, will its position be positive or negative 60 seconds after it passes the c

amera? If we multiply two positive numbers, is the result positive or negative?
Mathematics
1 answer:
dexar [7]3 years ago
7 0

Step-by-step explanation:

I can only answer part 2 of your question, "If we multiply two positive numbers, is the result positive or negative?". if you multiply 2 positive numbers your answer would be positive.

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You and your friend each start a car-washing service.
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5 i think because you cant get the exact amount he has
7 0
3 years ago
Solve the proportion: A printer can print 11 color pages in 2 minutes. How many color pages can the printer print in 7 minutes?
Ann [662]

Answer:

38.5

Step-by-step explanation:

7/2 = 3.5. 11×3.5= 38.5

6 0
3 years ago
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
What is the solution set of 2x2-3x=5
Kryger [21]

You use the quadratic formula:

2x {}^{2} - 3x = 5

2x {}^{2} - 3x - 5 = 0

x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2}  

x = \frac{3 + - \sqrt{49} }{4}  

x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4}  

x = \frac{5}{2} \: and \: x = - 1

7 0
3 years ago
Help asap idek what to do
daser333 [38]

So to solve this question, your goal is to find out how the way it is solved is not correct.

Your answer would be: On the third line, the student adds the 8 to both sides instead of subtracting. The way the initial equation is given is

y-(-8)=-6(x-2). After distributing the six, the student should make the 8 positive because subtracting a negative makes a positive. After solving, the equation should look like: y(+8)=-6x+12, so you would subtract the 8 from both sides instead of adding it, and solve from there.

5 0
3 years ago
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