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2 years ago

15

A fish tank that is 5 feet long by 3 feet wide by 3 feet high has a 2 foot cubic stone placed in the bottom how much water will

the tank hold

1 answer:

maks197457 [2]2 years ago

4 0

Answer:

37ft³

Step-by-step explanation:

Volume of the fish tank = length × width × height = 5ft × 3ft × 3ft = 45ft³

Volume of the cubic stone = length × length × length = (2ft)³ = 8ft³

Volume of the water = volume of the fish tank - volume of the cubic stone

i.e Volume of the water = 45ft³ - 8ft³ = 37ft³

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Double the quotient of 3 and 6

Hitman42 [59]

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This statement can be represented as $2*\frac{3}{6}$ , since the quotient is the answer to a division problem.

Divide. $2*\frac{1}{2}$

Multiply. $2*\frac{1}{2}=\frac{2}{1}*\frac{1}{2}=\frac{2*1}{1*2}=\frac{2}{2}$

Simplify. When the numerator and denominator are the same, the fraction is equal to 1. $\frac{2}{2}=1$

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2 years ago

Part A- Create a fourth degree polynomial in standard form. How do you know it is in standard form?

Alexeev081 [22]

Answer:

(a) $2x^4-9x^3+x^2+x-5$

Step-by-step explanation:

(a)The degree of a polynomial is the highest power of the unknown variable in the polynomial.

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An example of a fourth degree polynomial is: $2x^4-9x^3+x^2+x-5$

We know the polynomial above is in standard form because it is arranged in such a way that the powers of x keeps decreasing.

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change. This is illustrated by the two examples below:

$(2x^5-3x^2+x)+(3x^5-5x^2)\\=2x^5-3x^2+x+3x^5-5x^2\\=5x^5-8x^2+x\\$Similarly\\(2x^5-3x^2+x)-(3x^5-5x^2)\\=2x^5-3x^2+x-3x^5+5x^2\\=-x^5+2x^2+x$

The degrees do not change in the above operations. Only the number beside each variable changes. Therefore, the addition and subtraction of polynomials is closed.

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2 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

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Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

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Upon taking square root of both sides of our equation we will get,

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Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

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Upon substituting our given values we will get,

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Upon taking square root of both sides of our equation we will get,

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Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

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$\text{The length of the longer diagonal}=13.54$

Therefore, the length of longer diagonal is approximately 13.54 inches.

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2 years ago

Given: 2x^2+3x-5, g(x)=x+9. Find: (fg)(3)

Svetradugi [14.3K]

Answer:

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1 year ago

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2 years ago