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melisa1 [442]
3 years ago
12

Tina has a rectangular planter box. She decides that she needs more planting space so she triples the length, doubles the width

and leaves the height alone. If the original box is full of soil, how many times greater is the volume of soil in the new box?
Mathematics
1 answer:
kirill [66]3 years ago
4 0

Answer:

  6

Step-by-step explanation:

The volume is proportional to length, so tripling the length multiplies the volume by 3.

The volume is proportional to width, so doubling the width multiplies the volume by 2.

Tina's new planter box is 3×2 = 6 times the volume of the original.

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Margaret [11]

Answer:

B-4

Step-by-step explanation:

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2 years ago
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xy=c²==>y=c²/x
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2 years ago
The energy expended by a bird per day, E, depends on the time spent foraging for food per day, F hours. Foraging for a shorter t
Veronika [31]

Answer:

Therefore F=2.387 hours gives a minimum value of energy expenditure E.

Step-by-step explanation:

Given that,

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E=0.25 F+\frac{1.7}{F^2}

Differentiating with respect to F

E'=0.25 -\frac{3.4}{F^3}

Again differentiating with respect to F

E''=\frac{10.2}{F^4}

Now set E'=0

0.25 -\frac{3.4}{F^3}=0

\Rightarrow \frac{3.4}{F^3}=0.25

\Rightarrow F^3=\frac{3.4}{0.25}

\Rightarrow F=2.387

Now E''|_{F=2.387}=\frac{10.2}{2.387^4}>0

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Therefore F=2.387 hours that minimizes energy expenditure.

7 0
2 years ago
f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

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and

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2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

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