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REY [17]
3 years ago
10

Laura has two large cylindrical barrels of solution in her chemical laboratory. The first barrel contains 900 liters of solution

, and the second barrel contains 100 liters of solution.
Laura drains the first barrel at a rate of 36 liters per minute.

Laura fills the second barrel at a rate of 44 liters per minute.

Which inequality represents the minutes, m, when the second barrel contains a greater or equal amount of solution than the first barrel?
Mathematics
1 answer:
aleksley [76]3 years ago
6 0

Answer:

-36m + 900 < 44m + 100

Step-by-step explanation:

Given that:

The first barrel contains 900 liters of the solution; &

The second barrel contains 100 liters of solution.

So, Laura drains the first barrel at the rate of 36 liters/ min

Since Laura is draining the solution in the first barrel;

Then; we have - 36m + 900 litres

Also; Laura fills the second barrel at a rate of 44 liters per minute.

Since Laura is filling the barrel; we have: 44m + 100

Therefore; the inequality representing when the second barrel contains a greater or equal amount of solution than the first barrel can be expressed as:

- 36m + 900 < 44m + 100

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yanalaym [24]

Step-by-step explanation:

It is given that ∠SPQ≅∠SRQ

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles S P Q and S R Q are highlighted in red.

It is also given that SQ⎯⎯⎯⎯⎯

bisects ∠PSR

.

By the definition of angle bisector, ∠PSQ≅∠QSR

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are congruent and highlighted in red.

△PQS

and △RQS share a common side SQ⎯⎯⎯⎯⎯, and SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

by the Reflexive Property of Congruence.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Segment S Q is highlighted in red.

Two angles, ∠SPQ

and ∠PSQ, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △PQS are congruent to two angles, ∠SRQ and ∠QSR, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △RQS

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are highlighted in red. Angles S P Q and S R Q are highlighted in red. Side S Q is highlighted in blue.

So, △PQS≅△RQS

by the Angle-Angle-Side (AAS) Congruence Theorem.

PS⎯⎯⎯⎯⎯

and SR⎯⎯⎯⎯⎯ are corresponding sides of congruent triangles, △PQS and △RQS. So, PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

by CPCTC.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Sides S R and S P are congruent and highlighted in red.

Translate these six statements and reasons into a 2

-column proof,

1. ∠SPQ≅∠SRQ

(Given)

2. SQ⎯⎯⎯⎯⎯

bisects ∠PSR

. (Given)

3. ∠PSQ≅∠QSR

(Def. of ∠

bisect)

4. SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

(Reflex. Prop. of ≅

)

5. △PQS≅△RQS

(AAS Steps 1, 3, 4)

6. PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

(CPCTC)

There you go

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Answer: "Even after the fire, Brown was confident in his decision not to correct the mistake."

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ioda

Answer:

200 mL

Step-by-step explanation:

If x is the volume of solution B:

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30 + 0.20 x = 50 + 0.10 x

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Answer:

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Step-by-step explanation:

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Find the smallest positive integer n such that n(n+1)(n+2) is divisible by 247.
Butoxors [25]

Answer:

37 is the smallest positive integer n such that n(n+1)(n+2) is divisible by 247.

Step-by-step explanation:

First we will find the prime factors of 247:

247 = 13 x 19 (which are both prime).

So now we need to find a number (the smallest one) that is of the form (n)(n+1)(n+2) (the product of three consecutive numbers) and that is divisible by both 13 and 19 (and therefore divisible by 247)

Let's take a look at the multiples of 13: 13, 26, 39, 52...

Let's take a look at the multiples of 19: 19, 38, 57...

We can see that the first time we have two multiples close together are the 38 (for 19) and the 39 (for 17).

So, if our number has both 38 and 39 as factors, then it will be divisible by 247.

However, we need not two but three consecutive numbers, and since we want the number to be the smallest positive integer, we will add 37 (since our other choice would be to add 40 and that would make the number bigger) and thus our number is (37)(38)(39) or in other words (37)(37 + 1)(37 + 2) and therefore this is the smallest positive number such that n(n+1)(n+2) is divisible by 247.

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