Each day, the cafeteria serves more than 200 apples with lunch. Write an inequality that represents this situation. Let a represent the number of apples.

Answers

Answer 1

Answer:

The expression for the inequality is written as a > 200.

What is inequality?

When two expressions are connected by a sign like "not equal to," "greater than," or "less than," it is said to be inequitable. The inequality shows the greater than and less than relationship between variables and the numbers.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that each day, the cafeteria serves more than 200 apples with lunch.

The inequality will be written as,

a > 200

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Answer 2

Answer:
Here is the answer
a > 200

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William put $100 in a bank account last year. Now he has 1/12 more money in the account. Which expressions represent how much money William has in his bank account now? Choose all answers that are correct. a.100-(1/12*100) b.100+(1/12*100) c.100+(13/12*100) d.13/12*100
4 times the sum of 14 + 26

2X + 2 wht is the value of X

Answers

Ultimately, it is impossible to solve an equation without two sides. If it was 2X=2, the answer would be X=1. But if it was 2X+2=0, the answer would be X=-1. The information given is impossible to work with,

A piece of equipment is purchased for $18,000.00 and has a salvage value of $2,000.00.The estimated life is then years and the method of depreciation is straight-line. Shipping costs total $620.00 and installation costs are $590.00. The book value at the end of year 10 is....A.)$1,110.00

B)$2,110.00

C)$3,110.00

D)$2,000.04

HEEEELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP!!!! XD

Answers

Hello one of my favvvvv sons!
This is Accounting math, so congrats at being in college level math!
Now let's solve this! :)

The Book Value of an asset is the asset's cost minus the assets accumulated depreciation.
Asset's costs:
           Purchase of equipment = $18,000
           Shipping costs = $620.00
           Installation costs = $590.00

Total = $19,210.00

Assets accumulated depreciation:

depreciation per year=(Cost of equipment-Salvage Value) ÷ Estimated Useful Life = (18000 - 2000) ÷ 10 = $ 1,600 per anum (per year)

           Cost: $18,000
           Salvage Value: $ 2,000
           Estimated Useful Life: 10 years

With this data, we can make the following chart provided below.
Looking at the chart we can conclude that hence, in the 10th year, the accumulated depreciation value is $ 16,000.

So the Book Value at the end of year 10 is...
an asset is the assets cost - the assets accumulated depreciation.
$19,210.00 - $16,000 = $ 3,210

Hope I helped! If you have any other questions or would like further explanation feel free to Pm me! Have a great day!!! :)

How do I make these points (3,-5)(-5,3) into a point slope form equation that contains the given points

Answers

$General\ equation\ for\ line\ in\ slope\ intercept\ form:\n\ny=ax+b\n\nTo\ find\ a\ and\ b\ substitude\ points\ (3,-5),\ (-5,3)\ into\ equation\n\n \left \{ {{-5=3a+b}\ \ \ \ \atop {3=-5a+b\ |*-1}} \right.\n\n \left \{ {{-5=3a+b} \atop {-3=5a-b}} \right. \n+----\n\n-8=8a\ \ \ |:8\na=-1\n\n-5=3\cdot(-1)+b\n-5=-3+b\n-5+3=b\nb=-2\n\ny=-x-2$

Steps:
1. First you have to find the slope
3+5= 8 -5-3=-8 Your slope is -1
2. You plug in one of the points into the point slope form
Y1-Y2=m(x-x1)
Y-3=-1(x+5)
Y-3=-1x+-5
3. Then, you add three to each side
Y=-1x+ (-2)

Find the mean of this set of numbers.9, –6, –8, 11, –4, 10

A.
48

B.
2

C.
12

D.
8

Answers

b. 2

The mean is 12 divided by 6, or 2.

Find the greatest common factor of the following monomials: 20a3 and 8a2

Answers

GCF of given monomials are $4a^2$

Solution:

Given that we have to find the greatest common factor

Given monomials are:

$20a^3 \text{ and } 8a^2$

When we find all the factors of two or more numbers, and some factors are the same, then the largest of those common factors is the Greatest Common Factor

Let us first find the GCF of 20 and 8 and then find GCF of variables and then multiply them together

GCF of 20 and 8:

The factors of 8 are: 1, 2, 4, 8

The factors of 20 are: 1, 2, 4, 5, 10, 20

Then the greatest common factor is 4

$GCF\ of\ a^3 \text{ and } a^2\n\na^3 = a^2 * a\n\na^2 = a^2$