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Ira Lisetskai [31]
3 years ago
5

Confused from where I left off PLEASE help

Mathematics
2 answers:
Leviafan [203]3 years ago
4 0

Not sure the right equation so i did both....

#1 - √12m√15m=1m√2

Simplifies to:

13.416408m2=1.414214m

Let's solve your equation step-by-step.

13.416408m2=1.414214m

Step 1: Subtract 1.414214m from both sides.

13.416408m2−1.414214m=1.414214m−1.414214m

13.416408m2−1.414214m=0

Step 2: Factor left side of equation.

m(13.416408m−1.414214)=0

Step 3: Set factors equal to 0.

m=0 or 13.416408m−1.414214=0

m=0 or m=0.105409

Answer:  m=0 or m=0.105409


#2 - √12m√15m

Simplifies to:

13.416408m2

=13.416408*m2

=13.416408*(m*m)

=13.416408m2


Can I get a brainliest Please & Thank you...

Julli [10]3 years ago
3 0

I believe [1] and [2] are again placeholders, and not the numbers 1 and 2. Combine the square roots on the left side:

\sqrt{12m}\cdot\sqrt{15m}=\sqrt{12m\cdot15m}=\sqrt{180m^2}

Notice that

180=2^2\cdot3^2\cdot5

so we have

\sqrt{180m^2}=\sqrt{2^2\cdot3^2\cdot5\cdotm^2}=\sqrt{(2\cdot3\cdot m)^2\cdot5}=\sqrt{(2\cdot3\cdotm)^2}\cdot\sqrt5

=\sqrt{(6m)^2}\cdot\sqrt5=6m\sqrt5

(where \sqrt{m^2}=m only if m\ge0)

So I would imagine the solution is [1] = 6 and [2] = 5.

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A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo
katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

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Step-by-step explanation:

Let the dimensions of the box be x, y and z

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Therefore, Volume of the boxV=x^2z

Volume of the box=12 ft^3\\

Therefore, x^2z=12\\z=\frac{12}{x^2}

The material for the base costs \$0.17/ft^2, the material for the sides costs \$0.10/ft^2, and the material for the top costs \$0.13/ft^2.

Area of the base =x^2

Cost of the Base =\$0.17x^2

Area of the sides =4xz

Cost of the sides==\$0.10(4xz)

Area of the Top =x^2

Cost of the Base =\$0.13x^2

Total Cost, C(x,z) =0.17x^2+0.13x^2+0.10(4xz)

Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

To minimize C(x), we solve for the derivative and obtain its critical point

C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2

Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

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