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3 years ago

Which ordered pair is a solution of the equation? 3x−y=13

Mathematics

2 answers:

melisa1 [442]3 years ago

7 0

(0, -13), (1, -10), or (2, -7)

hope this helps <3

irakobra [83]3 years ago

4 0

Answer:

Both (6,5) (3,−4)

Step-by-step explanation:

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G = f에서 5 = g는 무엇입니까?

Scilla [17]

Answer:

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g = 5

f = 5

Step-by-step explanation:

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3 years ago

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The temperature decreases 3° every hour. c = temperature change Which equation accurately shows a decrease over 5 hours?

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3 years ago

What is the solution set for the equation 2|-3n| + 10 = 50 I

Aleks04 [339]

Answer:

$n=-\frac{20}{3}\quad \mathrm{or}\quad \:n=\frac{20}{3}$

Step-by-step explanation:

$2\left|-3n\right|+10=50\\\\\mathrm{Subtract\:}10\mathrm{\:from\:both\:sides}\\\\2\left|-3n\right|+10-10=50-10\\\\Simplify\\\\2\left|-3n\right|=40\\\\\mathrm{Divide\:both\:sides\:by\:}2\\\\\frac{2\left|-3n\right|}{2}=\frac{40}{2}\\\\Simplify\\\\\left|-3n\right|=20\\\\\mathrm{Apply\:absolute\:rule}:\\\quad \mathrm{If}\:|u|\:=\:a,\:a>0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a\\$

$-3n=-20\quad \mathrm{or}\quad \:-3n=20\\\\-3n=-20\quad :\quad n=\frac{20}{3}\\\\-3n=20\quad :\quad n=-\frac{20}{3}\\\\n=-\frac{20}{3}\quad \mathrm{or}\quad \:n=\frac{20}{3}$

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3 years ago

B=2.35+0.25x

natulia [17]

Answer:

The answer is b it's kinda hard to explain doe

Sorry!

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3 years ago

A storage box with a square base must have a volume of 90 cubic centimeters. the top and bottom cost $0.60 per square centimeter

Paul [167]

Answer-

For side length of 3.56 cm and height of 7.10 cm the cost will be minimum.

Solution-

Let us assume that,

x represents the length of the sides of the square base,

y represent the height.

Given the volume of the box is 90 cm³, so

$\Rightarrow V=90\\\\\Rightarrow x^2\times y=90\\\\\Rightarrow y=\dfrac{90}{x^2}$

As the top and bottom cost $0.60 per cm² and the sides cost $0.30 per cm². Total cost C will be,

$C=\text{cost for top and bottom}+\text{cost for rest 4 sides}\\\\=(2x^2\times 0.6)+(4xy\times 0.3)\\\\=(2x^2\times 0.6)+(4x\times \frac{90}{x^2}\times 0.3)\\\\=1.2x^2+ \dfrac{108}{x}$

Then,

$C'=\dfrac{d}{dx}(1.2x^2+ \dfrac{108}{x})=2.4x-\dfrac{108}\\\\C''=\dfrac{d^2}{dx^2}(1.2x^2+ \dfrac{108}{x})=2.4+\dfrac{216}{x^3}$

As C'' has all positive terms so, for every positive value of x (as length can't be negative), C'' is positive.

So, for minima C' = 0

$\Rightarrow 2.4x-\dfrac{108}{x^2}=0\\\\\Rightarrow 2.4x=\dfrac{108}{x^2}\\\\\Rightarrow x^3=\dfrac{108}{2.4}=45\\\\\Rightarrow x=3.56$

Then,

$y=\dfrac{90}{x^2}$

$y=\dfrac{90}{3.56^2}$

$y=7.10$

Therefore, for side length of 3.56 cm and height of 7.10 cm the cost will be minimum.

4 0

4 years ago

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