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Anna [14]
2 years ago
9

Abby feeds her dog 1 1/2 cups of food each day, and Braxton feeds his dog 4 1/2 cups of food each day. In one week, how many tim

es more dog food does Braxton’s dog eat when compared with Abby’s dog?
Braxton’s dog eats 5 times more food.
Braxton’s dog eats 3 times more food.
Braxton’s dog eats 4 times more food.
Braxton’s dog eats 2 times more food.
Mathematics
2 answers:
ElenaW [278]2 years ago
6 0
  • Abby's dog eats each day=1 1/2=3/2=1.5cups
  • Braxtons dog eats each day=4 1/2=9/2=4.5cups

Find ratio

\\ \tt\hookrightarrow \dfrac{4.5}{1.5}=3

Braxtons dog eats 3times more than Abbys

ddd [48]2 years ago
4 0

Solution:

<u>Note that:</u>

  • Abby's dog: 1.5 cups of food
  • Braxton's dog: 4.5 cups of food

Looking at the information we got, we can say that Braxton's dog eats more. Let's put them into ratio form (Braxton's dog:Abby's dog)

  • => (Braxton's dog:Abby's dog) = 4.5:1.5

<u>Now, put them into fractions.</u>

  • => 4.5:1.5 = 4.5/1.5 = 3

We can conclude that Braxton's dog eats 3 times more food.

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Answer:

Step-by-step explanation:

y = k*x                    This is the formula for a direct variation.

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20 = 4 * k                        Divide by 4

20/4 = 4k/4

k = 5

Answer: the constant of variation is 5

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(9x^2+8x)-(2x^2+3x)
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Answer:

d = \frac{7}{2}

Step-by-step explanation:

The nth term of an arithmetic sequence is

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where a₁ is the first term and d the common difference

Given a₉ = 16 , then

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Can someone help me
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Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

  (2x -12)/2 = arcsin(16/22×sin(24°))

  x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

  2x -12 = 0

  x -6 = 0 . . . . . divide by 2

  x = 6 . . . . . . . add 6

Then the range of values of x will be ...

  6 < x < 23.206

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* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

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4 0
3 years ago
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Answer:

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We can then rewrite the square root terms as follows:

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Next, we can use the property of square roots that says that the square root of a number is equal to the square root of each of its prime factors. This means that we can rewrite the square root term as follows:

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Since the square root of a number is the same as the number itself, we can simplify the expression further by removing the square root symbols from the prime numbers 2:

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Since any number divided by itself is equal to 1, we can simplify the expression one last time to get our final answer:

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