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Lapatulllka [165]
3 years ago
15

Shane and Abha earned a team badge that required their team to collect no less than 2000 cans for recycling. Abha collected 178

more cans than Shane did. Write an inequality to determine the number of cans, S, that Shane could have collected.
Mathematics
1 answer:
Katarina [22]3 years ago
4 0

Answer:

2x+178\geq 2000

Step-by-step explanation:

Let, the number of cans collected by Shane = x.

So, the number of cans collected by Abha = x + 178.

Since, at least 2000 cans are required to be collected.

Thus, we have the inequality,

Number of cans by Shane + Number of cans by Abha ≥ 2000.

i.e. x+(x+178)\geq 2000

i.e. 2x+178\geq 2000

Thus, the required inequality is 2x+178\geq 2000.

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The student council is selling tickets to a school dance. One teacher decides to pay $10 for his class, and the students in the
Lostsunrise [7]

Hi,

x + y = 350                    (equation 1)

3x + 2y = 950                 (equation 2)


Use Substitution Method

Isolate variable x in equation 1


x = 350 – y                     (equation 3)


Substitute equation 3 into equation 2


3(350 – y) + 2y = 950

1050 – 3y + 2y = 950

3y – 2y = 1050 – 950

y = 100


Substitute y = 100 into equation 1


x + 100 = 350

x = 250


Answer: 250 $3 tickets and 100 $2 tickets were sold.


8 0
3 years ago
An arithmetic sequence is given below.
Mashcka [7]
Given that a_1=24 and a_2=17, if a_n is an arithmetic sequence, then the common difference between successive terms is d=a_2-a_1=17-24=-7.

You then have

a_2=a_1+d
\implies a_3=a_2+d=a_1+2d
\implies a_4=a_3+d=a_1+3d
\implies \cdots\implies a_n=a_{n-1}+d=\cdots=a_1+(n-1)d

So the explicit formula for the nth term is

a_n=24-7(n-1)
3 0
3 years ago
A loan of $500 is to be paid with two payments of $300, one in 3 months, and another in 6 months. The compound interest charged
Alenkasestr [34]
The anual interest would be $200

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7 0
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nikklg [1K]
Maybe you should work on starting you're sentences.
3 0
3 years ago
In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th
WITCHER [35]

Answer:

a) 0.5198 computers per household

b) 0.01153 computers

Step-by-step explanation:

Given:

number of computers in a home,

q = 0.3458 ln x - 3.045 ;   10,000 ≤ x ≤ 125,000

here x is mean household income

mean income = $30,000

increasing rate, \frac{dx}{dt} = $1,000

Now,

a) computers per household are

since,

mean income of  $30,000 lies in the range of 10,000 ≤ x ≤ 125,000

thus,

q = 0.3458 ln(30,000) - 3.045

or

q = 0.5198 computers per household

b) Rate of increase in computers i.e \frac{dq}{dt}

\frac{dq}{dt} = \frac{d(0.3458 ln x - 3.045)}{dt}

or

\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0

on substituting the values, we get

\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000

or

= 0.01153 computers

6 0
3 years ago
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