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Zielflug [23.3K]
3 years ago
8

What is the answer to number 4 and how did you solve it?

Mathematics
1 answer:
blagie [28]3 years ago
3 0
This is just the perimeter so add all of the numbers up. 4.2 + 9.4 + 4.6 + 5.6 = 23.8.
 
Multiply 23.8 by 7.4 to convert it to meters. 23.8 X 7.4 =171.36
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She was charged 12 cents per kWh
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if rounded to the nearest dime what is the greatest amount of money that rounds to $105.40? what is the least amount of money th
Kamila [148]
The greatest amount of money would be $105.40.  The least amount would also be $105.40.  Because the zero is less than five the dollar amount would stay the same.
4 0
3 years ago
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Again, please help. i dont know any of this
Levart [38]

\textrm {Cylinder volume :}

\mathrm {Volume = \pi r^{2}{h}}

\mathrm {Volume = 3.14 \times (\frac{4}{2})^{2} \times 3}

\mathrm {Volume = 3.14 \times 12}

\mathrm {Volume = 37.68cm^{3}}

\textrm {Rectangular prism volume :}

\mathrm {Volume = length \times width \times height}

\mathrm {Volume = 6 \times 5 \times 7}

\mathrm {Volume = 30 \times 7}

\mathrm {Volume = 210cm^{3}}

\textrm {Total volume of composite figure :}

\textrm {Volume of cylinder + volume of rectangular prism}

\mathrm {37.68 + 210}

\mathrm {247.68cm^{3}}

4 0
2 years ago
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What is the directrix of the parabola? (y+4)2=4(x−2)
nika2105 [10]

Answer:

the answer is 1

Step-by-step explanation:

7 0
3 years ago
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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
4 years ago
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