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3 years ago

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2) Paul has two and two-thirds pounds of peanuts and a quarter of a pound of cashews . How many more pounds of cashews does he h

ave

1 answer:

Monica [59]3 years ago

4 0

Answer:

Step-by-step explanation:

2 2/3

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Find the volume of a cube that has a side length of:<br> 6x²y5

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3 years ago

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Maria had planned to walk 10 1/8 miles on wednesday, if she walked 7 2/8 miles in the morning how far would she need to walk in

Ugo [173]

Answer:

Distance Maria had to walk in the afternoon is $2\frac{7}{8}$

Step-by-step explanation:

Lets take that Maria walked $x$ miles in the afternoon.

Distance walked in the morning + distance walked in the afternoon = $10\frac{1}{8}$

⇒ $7\frac{2}{8} +x=10\frac{1}{8}$

$\frac{58}{8} +\frac{8x}{8} =\frac{81}{8}$

$\frac{58+8x}{8} =\frac{81}{8}$

$58+8x=81$

$8x=81-58\\$

$8x=23$

$x=\frac{23}{8}$

$x=2\frac{7}{8}$

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3 years ago

The circumference of the base of a cylinder is 24π mm. A similar cylinder has a base with circumference of 60π mm. The lateral a

earnstyle [38]

Given that the figures are similar polygons, the ratio of their ratios should be equal to the square of the ratio of their circumference. If we let x be the lateral area of the smaller cylinder then,
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The value of x from the equation is,
x = 33.6π
Thus, the area of the smaller cylinder is equal to 33.6π mm².

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3 years ago

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There were 5 girls and 22 boys in math express the number of girls as a fraction of the number of boys then express the fraction

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3 years ago

Evaluate the definite integral using the graph of f(x)<br> (Image included)

Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

$\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}$

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

$\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}$

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

$\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20$

Then by linearity, we have

$\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}$

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2 years ago