A shop produces 39 wet suits every 2 weeks. How long will it take to the shop to produce 429 suits?

Which shape must have opposite sides that are parallel and congruent, and diagonals that are perpendicular bisectors of each oth

n200080 [17]

Answer:

Rectangle or square

Step-by-step explanation:

The diagonals are perpendicular bisectors of each other. C) All four angles of the parallelogram are right angles. Congruent diagonals eliminates the rhombus. This leaves the rectangle or the square.

8 0

3 years ago

Read 2 more answers

In a board game, Rose gets x + 4 points when she lands on a green space, 7x – 1 points for landing on a blue space, and -4x poin

ra1l [238]

Answer:

5x-3

Step-by-step explanation:

x+4+7x-1-4x

5x-3

3 0

3 years ago

Read 2 more answers

a student spent 15 minutes painting a 2 foot by 3 foot bulletin board. to the nearest tenth of a minute, how long did it take th

Katyanochek1 [597]

Area of the board divided by the time

2 x 3 = 6ft^2

15/6= 2.5mins

8 0

2 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

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