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Alik [6]
3 years ago
14

The temperature in Miami, Florida is 22 degrees warmer than three times the temperature in Bangor, Maine. The temperature in Mia

mi is 82 degrees. Write an equation to determine the temperature in Bangor.
Mathematics
1 answer:
balandron [24]3 years ago
3 0

Answer:

See Explanation

Step-by-step explanation:

Let the temperatures of Miami and Bangor be M and B degrees respectively.

According to the given information:

M = 3B + 22... (required equation)

Plug M = 82

82 = 3B + 22

82 - 22 = 3B

60 = 3B

B = 60/3

B = 20 degrees

Thus, the temperature in Bangor is 20 degrees.

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A map has a scale of 2 inches = 15 miles.If two towns are 7 inches apart on the map,how far apart are they actually
seraphim [82]
1.) set up a w_k_u chart
2.) in the w column write in on top and mi in the bottom like this
w_K_U
in_
mi_
then in the k column write 2 for in and 15 for mi
W_K_U
in_2_
mi_15_
then in the U column, you write 7 for in and x for mi
W_K_U
in_2_7
mi_15_x
then cross multiply:  2x = 15(7)
multiply 15 times 7 which is 105
2x=105
divide 2x by 2
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3 years ago
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A taxi driver charges $3 for every kilometre driven, plus $4 per trip.
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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

(c)Height=3 feet

Step-by-step explanation:

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

The rectangular box has a square base.

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

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Isosceles triangle.
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