1. What is the value of x? Show all of your work.

1 answer:

4 0

Pythagorean Theorem: A^2 + B^2 = C^2
We know A and C, let's find B by doing it backwards 

C^2 - A^2 = B^2 
4^2 - 7 (square root of 7 squared is just 7) = B^2 
16 - 7 = B^2 
9 = B^2

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Find

shepuryov [24]

Answer:

The two odd numbers are 15 and 17

Step-by-step explanation:

Given

Let the odd numbers be represented with x and y

Let x be the greater number

$x + 5y = 92$

Required

Find x and y

Since x and y are consecutive odd numbers and x is greater, then

$x = y + 2$

Substitute y + 2 for x in $x + 5y = 92$

$y + 2 + 5y = 92$

Collect Like Terms

$y + 5y = 92 - 2$

$6y = 90$

Divide both sides by 6

$\frac{6y}{6} = \frac{90}{6}$

$y = \frac{90}{6}$

$y = 15$

Substitute 15 for y in $x = y + 2$

$x = 15 + 2$

$x = 17$

Hence; the two odd numbers are 15 and 17

4 0

3 years ago

Which is the best estimate for (6.3x10^-2)(9.9x10^-3) written in scientific notation?

ozzi

It should be 6.237*10^-4

8 0

3 years ago

Read 2 more answers

10 POINTS!!! FULL ANSWER WITH FULL STEP BY STEP SOLUTION PLEASE

Inga [223]

A) Profit is the difference between revenue an cost. The profit per widget is
m(x) = p(x) - c(x)
m(x) = 60x -3x^2 -(1800 - 183x)
m(x) = -3x^2 +243x -1800
Then the profit function for the company will be the excess of this per-widget profit multiplied by the number of widgets over the fixed costs.
P(x) = x×m(x) -50,000
P(x) = -3x^3 +243x^2 -1800x -50000

b) The marginal profit function is the derivative of the profit function.
P'(x) = -9x^2 +486x -1800

c) P'(40) = -9(40 -4)(40 -50) = 3240
Yes, more widgets should be built. The positive marginal profit indicates that building another widget will increase profit.

d) P'(50) = -9(50 -4)(50 -50) = 0
No, more widgets should not be built. The zero marginal profit indicates there is no profit to be made by building more widgets.

_____
On the face of it, this problem seems fairly straightforward, and the above "step-by-step" seems to give fairly reasonable answers. However, if you look at the function p(x), you find the "best price per widget" is negatve for more than 20 widgets. Similarly, the "cost per widget" is negative for more than 9.8 widgets. Thus, the only reason there is any profit at all for any number of widgets is that the negative costs are more negative than the negative revenue. This does not begin to model any real application of these ideas. It is yet another instance of failed math curriculum material.