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ehidna [41]
3 years ago
6

For the end-of-year party, Mt.Rise Middle School ordered 112 pizzas. There were eight fewer veggie pizzas than there were pepper

oni pizzas. There were three times as many combo pizzas as pepperoni pizzas. Use the 5-D process to define a variable and write an equation for this situation. Then determine how many of each kind of pizza were ordered.
Mathematics
1 answer:
rodikova [14]3 years ago
5 0
Let the number of pepperoni pizzas be x
pepperoni = x
veggie = x - 8    [<span>There were eight fewer veggie than pepperoni]
</span>combo = 3x       [T<span>here were three times as many combo as pepperoni]
</span>
Given that the total Pizza: 112
x + x - 8 + 3x = 112
5x - 8 = 112
5x = 112 + 8
5x = 120
x = 24

x= 24
x - 8 = 16
3x = 72

So there were 24 pepperoni pizza, 16 veggie pizza and 72 combo pizza

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Complete question :

On average, light emitted from the Sun takes 8 minutes 19 seconds to reach Earth. Using your approximated speed of light from part (b), as well as the equation distance = rate x time, calculate the approximate distance from the Earth to the Sun in meters. Write your answer in standard form and in scientific notation.

Answer:

1.497 * 10^8km

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Step-by-step explanation:

Given that:

Time taken = 8 minutes 19 seconds

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(3 * 10^8) * ((8 * 60) + 19)

(3 * 10^8) * 499

= 1497 * 10^8 m or 1497 * 10^5 km = 1.497 * 10^8km

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Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as \sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} }  ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\

Using these result in equation a we get

f(x) = \sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)

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Differentiating b on both sides with respect to x we get

f '(x) = f '(x)=\sqrt{2}  \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2  => x<π/4  => x∈[0,π/4]

2) π/4 + x > 3π/2  => x > 5π/4  => x∈[5π/4,2π]

from conditions 1 and 2 we have  x∈(0,π/4)∪(5π/4,2π)

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