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eimsori [14]
3 years ago
15

A bucket holds 10 potatoes. Hal removes and cleans 4 potatoes per minute.

Mathematics
1 answer:
nexus9112 [7]3 years ago
4 0

Answer:

an = 10 – 4(n – 1)

Step-by-step explaination

answer is an = 10 – 4(n – 1)

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How to solve for s with literal equations
olya-2409 [2.1K]
B = (1/6).π.x².s

1st Multiply both sides with 6 → 6B = (6).(1/6).π.x².s. Simplify by 6
6B = π.x².s. Now divide both sides by .π.x².
6B/(.π.x²)  =  .(π.x².s)/(.π.x²) After simplification you get:

6B/(.π.x²) = s →  s = 6B/(.π.x²)

8 0
3 years ago
What is the approximate area of rectangle GHIJ?<br> A: 902<br> B: 434<br> C: 672<br> D: 784
Julli [10]

Answer:

B. 434

Step-by-step explanation:

sorry for the late response! it’s 434. I worked out the problem & took the test, B is the correct answer.

4 0
3 years ago
1. Michael and 6 friends went to the movies. There, they each bought movie tickets and a bag of popcorn to share among them all.
Aleonysh [2.5K]

Answer:

The variable is t

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Step-by-step explanation:

Given:

Total cost = 7t + 4.50

The variable is t

Michael + 6 friends = 7 people

4.50 represent the cost of popcorn

t represent the cost of movie tickets

For instance,

If the cost of a movie tickets is $2

Then,

t = 2

Total cost = 7t + 4.50

= 7(2) + 4.50

= 14 + 4.50

= 18.50

Total cost = $18.50

5 0
3 years ago
Compare fractions using benchmarks
Vadim26 [7]
You can compare fractions by using the benchmark fraction of 1/2. To use this in order to make comparisons, you will need to look carefully at what the numerator and The denominators are in a fraction. If the top number, the numerator is less than half of the denominator, then the fraction would be less then the benchmark 1/2. If the numerator is more than half of the denominator then the fraction would be more than the benchmark 1/2. For example, 3/4 is more than 1/2 because the numerator of three is more than half of the denominator 4.
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3 years ago
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bearhunter [10]

Let x be the smallest of the consecutive integers summing to 2020. Then the largest integer in the sum is x+4039, so our answer is \boxed{4039}.

For completeness, though, let's solve for the value of x that satisfies the conditions of the problem. Recall that the sum of an arithmetic series with first term a_1, last term a_n, and n terms, is \frac{n(a_1+a_n)}{2}. Plugging in our known values gives us the equation \frac{4040(2x+4039)}{2}=2020, which simplifies to 2x+4039=1. Thus, x=-2019, so the 4040 consecutive integers are -2019,-2018,\dots,-1,0,1,\dots,2018,2019,2020. Notice how everything except the 2020 cancels out!

4 0
3 years ago
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