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3 years ago

I need this ASAP please! It’s on iready !!!!!

Mathematics

1 answer:

Alisiya [41]3 years ago

3 0

Answer:

18.7+6

Step-by-step explanation:

pick the first one

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5s-(100+2s =4s-2(10+s) how much boxes would they need to sell to get a equal amount each group must sell to have equal profits.

antoniya [11.8K]

5s-100-2s=4s-20-2s
3s-100. = 2s-20
+20. +20
3s-80. = 2s
-3s. -3s
-80. = -1s
-1 -1
80. = s

s=80

6 0

4 years ago

Please helpppppp meeeee I neeeed answerssssssss

MissTica

I've done the first four questions. Use this method to answer the rest. For an odd number of objects find the cost of one and multiply by how many you are asked to find.

8 0

3 years ago

The measure of angle KMN = ?? degrees

Olegator [25]

∠KMN = (1/2)*arc KN . . . . . . this relationship is true of any inscribed angle. (It is a useful relationship to remember.)
  = 95°/2
  = 47.5°

5 0

3 years ago

A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro

Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

6 0

4 years ago

BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Misha Larkins [42]

146

hope this helps :D

3 0

3 years ago