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RideAnS [48]
3 years ago
9

49 oz to ___ lb ____oz

Mathematics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

3lb .6 ounces should be correct

Step-by-step explanation:

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To convert a distance of 13,000 feet to miles, which ratio could you multiply<br> by?
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Answer:

1 mile

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5280 feet

Step-by-step explanation:

There are 5,280 feet in one mile, so to convert it, multiply by ^

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Given 2x⁵-6x³-4x²-2x+4 =0 solve for x
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Step-by-step explanation:

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2 years ago
Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.
irina1246 [14]
Lagrange multipliers:

L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)

L_x=y^2z^2+\lambda=0
L_y=2xyz^2+\lambda=0
L_z=2xy^2z+\lambda=0
L_\lambda=x+y+z-5=0

\lambda=-y^2z^2=-2xyz^2=-2xy^2z

-y^2z^2=-2xyz^2\implies y=2x (if y,z\neq0)

-y^2z^2=-2xy^2z\implies z=2x (if y,z\neq0)

-2xyz^2=-2xy^2z\implies z=y (if x,y,z\neq0)

In the first octant, we assume x,y,z>0, so we can ignore the caveats above. Now,

x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of f(1,2,2)=16.

We also need to check the boundary of the region, i.e. the intersection of x+y+z=5 with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force f(x,y,z)=0, so the point we found is the only extremum.
4 0
3 years ago
Mr. Blue drove from Allston to Brockton, a distance of 105 miles. On his way back, he increased his speed by 10 mph. If the jour
andrew-mc [135]

Answer:

His original speed was 60 mph.

Step-by-step explanation:

The average speed formula is shown below:

speed = distance / time

For the first leg of the trip Mr. Blue drove at a speed of "x" mph, for a distance of 105 miles, therefore the time that took to complete the trip is:

time 1 = distance / speed

time 1 = 105 / x h

On the second leg of the trip Mr. Blue drove faster at a speed of "x + 10" mph, so the time it took him to complete the trip was 15 minutes less, therfore:

time 2 = time1 - (15/60) = (105/x) - 0.25 = (105 - 0.25*x)/x h

Applyint the time and speed from the second leg to the average speed formula we have:

x + 10 = {105/[(105 - 0.25*x)/x]}

x + 10 = 105*x/(105 - 0.25*x)

(x + 10)*(105 - 0.25*x) = 105*x

105*x + 1050 -0.25*x² - 2.5*x = 105*x

-0.25*x² -2.5*x + 1050 = 0 /(-0.25)

x² + 10*x - 4200 = 0

x1 = [-(10) + sqrt((10)² - 4*(1)*(-4200))]/2 = 60

x2 = [-(10) - sqrt((10)² - 4*(1)*(-4200))]/2 = -70

Since his speed couldn't be negative the only possible value was 60 mph.

5 0
3 years ago
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Please answer as soon as possible!!!<br> Will give brainlyest!!!!!!!!!!!
8_murik_8 [283]
The correct answer is C.

3 0
3 years ago
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