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nordsb [41]
4 years ago
11

#1 $11.49 for 3 packages and fid the unit rate.round to the nearest hundredth if necessary

Mathematics
2 answers:
fomenos4 years ago
8 0
Solve:-

$11.49 ÷ 3 = 3.83
$3.83 per package.

Unit Rate:- <span>$3.83 per package.</span>
mote1985 [20]4 years ago
3 0
If you would like to find the unit rate, you can calculate this using the following steps:

$11.49 ... 3 packages
$x = ? ... 1 package

11.49 * 1 = 3 * x
11.49 = 3 * x         /3
x = 11.49 / 3
x = $3.83

The correct result would be $3.83.
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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
Geometry question shown above. Please help.
zhannawk [14.2K]
For question 4 its (C)
6 0
3 years ago
Find the angle of depression from the top of a lighthouse, 260 feet above water to a ship that's 270 feet offshore?
expeople1 [14]

Answer: OPTION C.

Step-by-step explanation:

Observe the triangle ABC attached.

Notice that the angle of depression is represented with \alpha.

Knowing that the top of a lighthouse is 260 feet above water and the ship is 270 feet offshore, you can find the value of  \alpha by using arctangent:

\alpha= arctan(\frac{opposite}{adjacent})

In this case you can identify that:

opposite=260\\adjacent=270

Therefore, substuting values into  \alpha= arctan(\frac{opposite}{adjacent}), you get that the angle of depression is:

 \alpha= arctan(\frac{260}{270})\\\\\alpha=43.92\°

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bp%7D%7B5%7D" id="TexFormula1" title="\frac{p}{5}" alt="\frac{p}{5}" align="absmiddle
tatyana61 [14]

Answer:

p ≥ -10

Step-by-step explanation:

first, subtract one from each side to get:

p/5 ≥ -2

lastly, multiply each side by 5

p ≥ -10

6 0
2 years ago
Factor the polynomial. 48w^7 30w^4 a) 6w^4(8w^3 5) b) 6w^3(8w^4 5w) c) w^4(48w^3 30) d) 6(8w^7 5w^4)
Oliga [24]
48w^7 + 30w^4 = 6w^4(8w^3 + 5)
8 0
3 years ago
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