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Aleksandr-060686 [28]
3 years ago
6

Can someone help me turn these two word problems into algebraic expressions?

Mathematics
1 answer:
Dima020 [189]3 years ago
5 0
1. N=x+3
    S=2(x+3), or Sam=2x+6
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Which statement best describes how to find the distance between-5 and 15​
Alex787 [66]

Answer:  A

Step-by-step explanation:

4 0
4 years ago
Find the area of the region defined by the region defined by the inequality 2|x| + 3|y-1| ≤ 6
EleoNora [17]

If x and y-1 have the same sign, then either

x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9

or

x

If x and y-1 have opposite sign, then

x>0,y

or

x1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}

The area of R is given by the double integral

\displaystyle \iint_R dx\,dy

To compute the area, change the variables to

\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}

with determinant \det(J) = -\frac1{12}. Then the integral transforms to

\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}.

5 0
2 years ago
List and describe the characteristics of a perfectly competitive market.
stepladder [879]
The three primary characteristics of perfect competition are (1) no company holds a substantial market share, (2) the industry output is standardized, and (3) there is freedom of entry and exit. The efficient market equilibrium in a perfect competition is where marginal revenue equals marginal cost.
3 0
3 years ago
Q4:<br> John runs 15 miles in 3 hours. How many miles can John run per hour?
denis-greek [22]

Answer:

5 per hour

Step-by-step explanation:

15 miles the total amount he ran divided by 3 the total amount of time

15/3=5 so John ran 5 mph

8 0
3 years ago
Read 2 more answers
How do I calculate the distance between these two lines?<br><br> Y=-2/3x - 1/2 and Y=-2/3 + 1/5
motikmotik
You would have to graph it first
The formula for distance is (y2-y1)/(x2-x1)
3 0
3 years ago
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