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dlinn [17]
2 years ago
10

Herschel uses an app on his smartphone to keep track of his daily calories from meals. One day his calories from breakfast were

more than his calories from​ lunch, and his calories from dinner were less than twice his calories from lunch. If his total caloric intake from meals was ​, determine his calories for each meal.
Mathematics
1 answer:
Nikitich [7]2 years ago
6 0

Answer:

let the number of calories from lunch be called L. As such, breakfast is then L + 128, and dinner is 2L - 400. We can then sum the three meals and equate it to the total caloric intake, the known value of 1932.

  So: 1932 = L + L + 128 + 2L - 400 = 4L - 272.

  Lunch = 551

Breakfast = 551 + 128 = 679

Dinner = 2*551 - 400 = 702

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The positive difference between the two roots of the quadratic equation $3x^2 - 7x - 8 = 0$ can be written as $\frac{\sqrt{m}}{n
Sergio [31]

3x² - 7x - 8 = 0

We're asked about square roots so we won't try to factor; we'll go right for the quadratic formula,

x = ( 7 ± √(7² - 4(3)(-8))    )/(2(3)) = (7 ± √(49+96))/6 = 7/6 ± √145/6

145 = 5×29, so no square factors.  The positive difference is

d =  (7/6 + √145/6) - (7/6 - √145/6) = 2√145/6 = √145/3

so m=145, n=3 for a sum of

Answer: 148

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2 years ago
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For the dilation, DO, K = (10, 0) → (5, 0), the scale factor is equal to _____.
eduard
Scale factor is equal to 1/2, since the point went from 10 to 5
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2 years ago
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Pat's Pizza made 21 cheese pizzas, 14 veggie pizzas, 23 pepperoni pizzas, and 14 sausage pizzas yesterday. Based on this data, w
Charra [1.4K]

Answer:

I think it is maybe 19% but Idk.

Because 14 would seem to low without work and the other are pretty high so im just guessing.

3 0
2 years ago
The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

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Question 1.

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0.5x+y<20

y≥2x

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