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Sonja [21]
3 years ago
7

Quentin is one third the age of his aunt. And 20% older than his sister. If Quentin's sister is 15 years old, how old is his aun

t?
Mathematics
2 answers:
s2008m [1.1K]3 years ago
7 0

Answer:

54

Step-by-step explanation:

Q is 20% older than his sister so 15+20%=18 and 18=1/3 aunts age

So 18x3=54

anastassius [24]3 years ago
5 0

Answer:

54

Step-by-step explanation:

20% of 15 is 3 so Quentin is 18 years old

If Quentin is one thrid his Aunt's age that means his aunt is three times older.

18 * 3 = 54

So his Aunt is 54 years old

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Over what interval will the immediate value theorem apply
koban [17]

Answer:

Any [a,b] that does NOT include the x-value 3 in it.

Either an [a,b] entirely to the left of 3, or

an  [a,b] entirely to the right of 3

Step-by-step explanation:

The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.

Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.

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Plsss help no one answered the other times I put this question
svetlana [45]

Answer:

a) <em>The equation</em> (10s + 8w) <em>represents </em><em>the </em><em>calories </em><em>Bridget </em><em>ate </em><em>on </em><em>Monday </em><em>and </em><em>the </em><em>equation</em> (20s + w) <em>represents </em><em>the </em><em>calories</em><em> </em><em>she </em><em>ate</em><em> </em><em>the </em><em>next </em><em>day.</em>

<em>b)</em><em> </em><em>The </em><em>number </em><em>of </em><em>calories</em><em> </em><em>in </em><em>each </em><em>strawberry</em><em> </em><em>is </em>4 <em>and </em><em>the </em><em>number </em><em>of </em><em>calories </em><em>in </em><em>each </em><em>vanilla</em><em> </em><em>wafer</em><em> cookie</em><em> </em><em>is </em>19. The solution is s= 4 and w = 19.

Step-by-step explanation:

For part A, Bridget ate 10 strawberries and 8 vanilla wafer cookies on Monday. Since the the number of calories in a strawberry is <em>s</em> and the number of calories in a vanilla wafer cookie is <em>w </em>, the number of calories Bridget ate on Monday is <em>10s + 8w</em><em>.</em><em> </em>The next day, Bridget ate 20 strawberries and 1 vanilla wafer cookie. Hence, the number of calories Bridget ate on the next day is 20s<em> + w</em>.

For part B,

we will create two different simultaneous equations.

Equation 1: 10s + 8w = 192

Equation 2: 20s + w = 99

We need to find one of the terms first to solve the other term. For this case, I will solve for w first.

Multiply the first equation by 2.

Equation 3: 20s + 16w = 192*2 = 384.

Now, subtract equation 2 from this new equation.

Equation 4:

(20s + 16w) - (20s + w) = 384 - 99

20s + 16w - 20s - w = 285

15w = 285

This leaves only w left and we can solve w.

w = 285 / 15 = 19

Now, we can solve for s using equation 2.

20s + 19 = 99

20s = 99-19 = 80

Hence,

s = 80/20 = 4

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Answer:

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Step-by-step explanation:

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