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BartSMP [9]
2 years ago
7

Iris has never had a pet before, but today she took in a stray cat named Luna. For Iris, kitty litter is most likely what type o

f purchase?
Mathematics
1 answer:
erik [133]2 years ago
5 0
the answer is Minor New Purchase Decision. <span>It refers to a decision, in which the purchase of something totally new for the consumer is involved, but the purchase is not important in terms of need and / or money</span>
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Which expressions are equivalent to the one below? Check all that apply.
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5^3(5^x)  the rule is a^b(a^c)=a^(b+c)

5^(3+x)
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3 years ago
Which point is located on quadrant IV
pychu [463]

It's the letter D hope it helps

8 0
2 years ago
Greg runs 3 miles in 28 minutes. At the same rate, how many miles would he run in 42 minutes?
Paladinen [302]

Answer:

4.5

 miles

Step-by-step explanation:

Well, he runs

3

miles in

28

minutes. Also, see that

42

=

28

⋅

1.5

.

So, his journey would just be running

1.5

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So, he runs a total of

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7 0
3 years ago
Read 2 more answers
A 10 pound bag of grapes costs $14.00. How much is it for 1 pound of grapes?
patriot [66]

Answer:

1.40

Step-by-step explanation:

I feel if you divide $14.00 by 10 you would get 1.40, so  I say this should be the right answer.

I apologize if I am incorrect.

3 0
2 years ago
Read 2 more answers
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

(c)Height=3 feet

7 0
3 years ago
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