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faltersainse [42]

1 year ago

14

C. If one cubic foot contains about 7.5 gallons, how many gallons of water are in the pool? (3 points) I Water: gallons 3. A rec

tangular cake pan is 13 inches by 9 inches by 2 inches. A round cake pan has a diameter of 8 inches and a height of 2 inches. Which will hold more batter, one rectangular pan or two round pans? (5 points)

1 answer:

Dominik [7]1 year ago

4 0

Answer:

<u>One Rectangular Pan</u>

Step-by-step explanation:

Part C cannot be answered since the dimensions of the pool are not given. Please check the question.

The cake pans may be compared by calculating the volume of each shape:

Rectangular pan: Volume = Length x Width x Height

Volume Rectangle = (13")*(9")*(2") = <u>234 in^3</u>

Round pan: Volume = $\pi r^{2} h$

Volume Round = (3.14)*(4")^2(2") = 100.5 in^3

Two round pans will hold 2*(100.5 in^3) or 201 in^3 of batter

One rectangular pan holds 234 in^3.

Go with the rectangular pan if you want more cake.

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inn [45]

Answer:

≈ 15.08 units²

Step-by-step explanation:

The area (A) of the sector is calculated as

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= π × 6² × $\frac{48}{360}$

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2 years ago

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Step-by-step explanation:

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Because they said 11% of remaining salary, we have to do:

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1 year ago

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What two rational expressions sum to 2x+3/x^2-5x+4

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Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

$5 = -3A$

$\frac{5}{-3} = \frac{-3A}{-3}$

$A = -\frac{5}{3}$

Step 4: Plug in the values of A and B into the original equation in step 2

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$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

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