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qaws [65]
3 years ago
9

Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that cont

ain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

320 bags

Step-by-step explanation:

<h3>Introduction</h3>

Let's first assign some literals, to simplify the problem. The goal is to set everything up, in order to only use one symbol.

p: number of bags with only peanuts.

a: number of bags with only almonds.

r: number of bags with only raisins.

x: number of bags with only raisins and peanuts.

Now, the problem establish 3 useful equations. We can find equations equivalences for the next sentences.

<em>"The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts"  </em>is equivalent to r = 10p.

<em>"The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts" </em>is equivalent to a= 20x.

<em>"The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds" </em>is equivalent to p = \frac{1}{5} a.

<h3>Now, let's set every variable in function of x</h3>

We already know that a = 20x.  And because of that, we also know that

p = \frac{1}{5}a = \frac{1}{5}(20x) = 4x

and to conclude this stage of the problem, we also know that r = 10p =10(4x) = 40x

<h3>¡Let's draw it!</h3>

As there are only 3 items, it is possible to use a Venn diagram. As we can see in the diagram, the entire quantity of bags is going to be

210 + 4x + x + 40x = 210 + 45x

But, we also know that there are 435 bags, then we only have to solve the equation:

210 + 45x = 435

45x = 435 - 210

45x = 225

x = 225/45

x = 5

<h3>Conclude </h3>

Substituting x = 5 we get

a = 20 x = 20(5) = 100

p = 4 x = 4(5) = 20

r = 40 x = 40(5) = 200

Finally ans = 100 + 20 + 200 = 320

<h3 />

<h2 />

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3 years ago
Janice is 7 years older than Tam. Complete the table, and then graph this situation.
qwelly [4]

Answer:

The table in the exercise can be completed with the next results:

  • x          y
  • 2         <u>9</u>
  • 4         <u>11</u>
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Step-by-step explanation:

As in the exercise Janice is 7 years older than Tam, to obtain the result in the table, you must add 7 to each age in the column x, with this we can make the next formula:

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Remember that x is Tam's age, and y is Janice's age, so, you must replace the x variable in each case to obtain the result to y:

When x is 2:

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When x is 4:

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When x is 8:

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At last, <u>to obtain the graph you can use the formula made: y = x + 7</u>, and you'll obtain a graph like the attached picture, <em>where each time x obtain a unit, the y variable obtain a unit too maintaining the diference of 7</em>.

8 0
3 years ago
Can you answer these 3 questions for me plz
finlep [7]
A.
1hr \: 15min = 75min

c = 0.22t = 0.22 \times 75 = 16.5

cost is $16.50

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26.40 = 0.22 \times t

rearrange to solve for t:

t = \frac{26.40}{0.22} = 120min

C. This is the same as part B, only the rate changed

t = \frac{26.40}{0.25} = 105.6min

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the answer is 105 minutes.
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3 years ago
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6 0
3 years ago
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In a G.P the 3rd term is 4 times the 1st term and the sum of the 2nd term and the 4th term is 30.find the common ratio,find the
STALIN [3.7K]

Answer:

the common ratio is either 2 or -2.

the sum of the first 7 terms is then either 765 or 255

Step-by-step explanation:

a geometric sequence or series of progression (these are the most common names for the same thing) means that every new term of the sequence is created by multiplying the previous term by a constant factor which is called the common ratio.

so,

a1

a2 = a1×f

a3 = a2×f = a1×f²

a4 = a3×f = a1×f³

the problem description here tells us

a3 = 4×a1

and from above we know a3 = a1×f².

so, f² = 4

and therefore the common ratio = f = 2 or -2 (we need to keep that in mind).

again, the problem description tells us

a2 + a4 = 30

a1×f + a1×f³ = 30

for f = 2

a1×2 + a1×2³ = 30

2a1 + 8a1 = 30

10a1 = 30

a1 = 3

for f = -2

a1×-2 + a1×(-2)³ = 30

-10a1 = 30

a1 = -3

the sum of the first n terms of a geometric sequence is

sn = a1×(1 - f^(n+1))/(1-f) for f <>1

so, for f = 2

s7 = 3×(1 - 2⁸)/(1-2) = 3×-255/-1 = 3×255 = 765

for f = -2

s7 = -3×(1 - (-2)⁸)/(1 - -2) = -3×(1-256)/3 = -3×-255/3 =

= -1×-255 = 255

4 0
2 years ago
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