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Lena [83]
2 years ago
13

A teacher plans to buy air fresheners that cost $3.50 each. If the teacher can spend no more than $10 on the air fresheners and

has a $3 discount coupon, which inequality represents the number of air fresheners, x, the teacher can buy?
Mathematics
1 answer:
iris [78.8K]2 years ago
8 0

Answer:

Step-by-step explanation:

We need to create an equation to represent how many air fresheners the teacher can buy.

Teacher has $10 and a $3 discount coupon which would give her an extra $3 to spend. Each air freshener is $3.50.

x = number of air fresheners

($10 + $3) / $3.50 = x        can be written as $13/$3.50 = x

We add $10 and $3 because that gives us the total amount she has to spend.

We divide by $3.50 because that is the cost of each air freshener. The result is the number of air fresheners she can buy.

In this case.

$13/$3.50 = x

3.71 = x   BUT we can't buy part of an air freshener, so she can purchase 3 air fresheners.

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A welder drops a piece of red-hot steel on the floor. The initial temperature of the steel is 2,500 degrees Fahrenheit. The ambi
sergey [27]

Answer: After 18.05 minutes, the temperature of steel becomes 100 degrees.

Step-by-step explanation:

Since we have given that

Initial temperature = 2500

At t = 0,

we get that

f(t)=Ce^{-kt}+80\\\\2500=C+80\\\\2500-80=C\\\\2420=C

After 2 minutes, the temperature of the steel is 1500 degrees.

so, it becomes,

1500=2420e^{-2k}+80\\\\1500-80=2420e^{-2k}\\\\\dfrac{1420}{2420}=e^{-2k}\\\\0.587=e^{-2k}\\\\\ln 0.587=-2k\\\\-0.533=-2k\\\\k=\dfrac{0.533}{2}\\\\k=0.266

So, We need to find the number of minutes when the temperature of steel would be 100 degrees.

So, it becomes,

100=2420e^{-0.266t}+80\\\\100-80=2420e^{-0.266t}\\\\20=2420e^{-0.266t}\\\\\dfrac{20}{2420}=e^{-0.266t}\\\\\ln \dfrac{20}{2420}=-0.266t\\\\-4.8=-0.266t\\\\t=\dfrac{4.8}{0.266}\\\\t=18.05

Hence, after 18.05 minutes, the temperature of steel becomes 100 degrees.

7 0
2 years ago
Which equation has the components of 0 = x2 – 9x – 20 inserted into the quadratic formula correctly? x = x = x = x =
Ber [7]
To solve the quadratic equation given by 0=x^2-9x-20, we use the quadratic formula given by:
x=[-b+\- sqrt(b^2-4ac)]/(2a)
where,
a=1,b=-9,c=-20
thus substituting the above values into our formula we get:
x=[9+\-sqrt(9^2-4(-20*1))/(2*1)
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6 0
2 years ago
Read 2 more answers
The diagram shows a pentagon. It has one line of symmetry.
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Answer:

A) 10x + 8 = 21

B) x = 1.3

Step-by-step explanation:

2(2x + 3) + 2(x + 1) + 4x = 21

4x + 6 + 2x + 2 + 4x = 21

10x + 8 = 21

10x = 13

x = 1.3 cm

7 0
2 years ago
Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈
uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
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