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Ray Of Light [21]

2 years ago

12

A coin weighs 4/15 ounces. A different coin weighs 1/15 ounces. How much do the two coins weigh together? Write your answer as a

fraction in simplest form.

1 answer:

olchik [2.2K]2 years ago

7 0

Answer:

1/3

Step-by-step explanation:

4/15 + 1/15 = 5/15

simplified its 1/3

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ahrayia [7]

Answer:

yes, 2 gallons

Step-by-step explanation:

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2 years ago

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Bingel [31]

Can’t help you with this, we need either the picture of the graph or the formula for the graph

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3 years ago

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WILL MARK BRIANLIEST!

gayaneshka [121]

Answer:

(4, - 2 )

Step-by-step explanation:

Given the 2 equations

x + y = 2 → (1)

x - y = 6 → (2)

Adding the 2 equations term by term will eliminate the term in y, that is

(x + x) + (y - y) = (2 + 6)

2x = 8 ( divide both sides by 2 )

x = 4

Substitute x = 4 in either of the 2 equations and solve for y

Substituting in (1)

4 + y = 2 ( subtract 4 from both sides )

y = - 2

Solution is (4, - 2 )

7 0

3 years ago

By what percentage must the diameter of a circle be increased to increase its area by 50%?

katovenus [111]

The circle increase in area will be by a factor of 1.5
So corresponding increase in the diameter will be a factor of sqrt 1.5 = 1.2247

Answer is 22.47 %

5 0

3 years ago

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A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t

solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by

V = $V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x$

First, we differentiate V(x) with respect to x, to get,

$\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450$

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0$

Solving, with the help of quadratic formula, we get,

$x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}$ ,

Again differentiation V(x), with respect to x, we get,

$\frac{d^2(V(x))}{dx^2} = 24x - 180$

At x =

$\displaystyle\frac{5(3-\sqrt{3})}{2}$ ,

$\frac{d^2(V(x))}{dx^2} < 0$

Thus, by double derivative test, the maxima occurs at

x = $\displaystyle\frac{5(3-\sqrt{3})}{2}$ for V(x).

Thus, largest volume the box can have occurs when $x = \displaystyle\frac{5(3-\sqrt{3})}{2}}$ .

Maximum volume =

$V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}$

8 0

3 years ago