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2 years ago

2x+3y=15 x+y=6 find values for x and y

Mathematics

1 answer:

MatroZZZ [7]2 years ago

8 0

Answer:

(x, y) = (3, 3)

Step-by-step explanation:

Subtract double the second equation from the first.

(2x +3y) -2(x +y) = (15) -2(6)

y = 3 . . . . . . . . simplify

Use the second equation to find x.

x = 6 -y = 6 -3 = 3

The values of x and y are both 3.

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The play director spent 190190190190 hours preparing for a play. That time included attending 35353535 rehearsals that took vary

Genrish500 [490]

Answer:

The equation above represents the total time the play director spent preparing for a play.

Step-by-step explanation:

The time spent by the play director for preparing for a play is, 190 hours.

Of these 190 hours, the director spent varying amounts of time attending 35 rehearsals for the play.

Let the varying amounts of time be denoted by, x.

The director also spent 3/4th of an hour, i.e. 45 minutes, on other responsibilities related to the play.

The equation provided is:

$35x+\frac{3}{4}=190$

The equation above represents the total time the play director spent preparing for a play.

7 0

3 years ago

25 + 14 = blank + 25

sleet_krkn [62]

Answer:

Step-by-step explanation:

25 + 14 = 14 + 25

39 = 39

6 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$