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3 years ago

6

Maureen used 0.9 meters of fabric to make a pillow. How many more pillows of this size can Maureen make with 12.75 meters of add

itional fabric?

1 answer:

Basile [38]3 years ago

4 0

Answer:

14

Step-by-step explanation:

12.75m/0.9m = 14

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Solve for x.<br> 3(2x + 5) = 33

Umnica [9.8K]

Answer:

x = 3

Step-by-step explanation:

3(2x + 5) = 33

3 * 2x + 3 * 5 = 33

6x + 15 = 33

6x = 33 - 15

x = 18 / 6

x = 3

6 0

3 years ago

Homework Progress 23 / 39 Marks Work out 83 +52

antoniya [11.8K]

Answer:

2 1/25 or 2.04

Step-by-step explanation:

8 ^ 1/3 + 5 ^ (-2)

Rewriting

(2^3) ^ 1/3 + 1/5 ^2

2 + 1/25

2 1/25

2.04

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3 years ago

Read 2 more answers

Expand the following expression

masya89 [10]

Answer:

$\frac{5}{4} \times 4x + \frac{5}{4} \times \frac{3}{4} = 5x + \frac{15}{16}$

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2 years ago

John walks m miles each week. Marie’s distance can be found by using the expression : 3m + ½

KiRa [710]

Answer:

21.5 Miles or 21 ½ Miles

Step-by-step explanation:

Plug in 7 as m to 3m + ½.

Then, I get 3(7) + ½.

That concludes to 21 + ½

Add the ".5" or "½" at the end

So, answer is you get is 21.5 of 21 ½

7 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 <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

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

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

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

3 years ago