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astraxan [27]
3 years ago
11

Standerd form example

Mathematics
2 answers:
PIT_PIT [208]3 years ago
7 0
Rewrite y=1/2x+4
we know that standard form equations cannot contain fractions. therefore, let's first get rid of the fractions. we can multiply by the Least Common Denominator (LCD) to get rid of the fraction.
y=1/2x+4

2[y=1,2x+4]          multiply all terms by 2.

2y=x+8                 distribute the 2 to all terms.

-x+2y=x-x+8           subtract x form both sides in order to x and y on the same                                      side of the equation.

-x+2y=8

-1[-x+2y=8]                  multiply all terms by -1 to make the lead coefficient                                                 positive.

x-2y=-8                               the equation written in standard form.

slega [8]3 years ago
3 0
3951 is an example of a standard form. 

What standard form means is the number is written in numerical form. 

More examples: 5269, 95862, 125634, etc. 

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Simplify<br> x2 + 5x + 6/<br> X + 2
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Answer:

x+3

Step-by-step explanation:

combine like terms then simple math

6 0
2 years ago
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Dale tried to estimate the area of his backyard garden. He guessed it was 12 square meters. When he measured the dimensions late
Elena-2011 [213]

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5 0
3 years ago
Factor completely 20x^3y + 30x^2y^2.
weqwewe [10]

Answer:

10x^2 y(2x + 3y)

Step-by-step explanation:

20x^3 y + 30x^2 y^2.

Factor 10x^2y out of 20x^3y.

10x^2 y (2x) + 30x^2 y^2  

Factor 10x^2y out of 30x^2y^2.

10x^2 y (2x) + 10x^2 y (3y)  

Factor 10x^2y out of 10x^2 y (2x) + 10x^2 y (3y).

10x^2 y(2x + 3y)

you factor out 10x^2y from both side which you will then get 10x^2 y (2x) + 10x^2 y (3y)  than you factor out 10x^2y again and get 10x^2 y(2x + 3y)  your third option

6 0
3 years ago
How many ways can the letters in the word balloon be arranged? 210 1,260 2,520 5,040
Lapatulllka [165]

Answer:

The number of ways this can be done is 1,260 ways

Step-by-step explanation:

In this question, we are asked to calculate the number of ways in which the letters of the word balloon can be arranged.

To do this, we take into consideration those letters that are repeated and the number of times repeated. The letters are l and o and are repeated two times each.

The number of ways = 7!/2!2! = 5040/4 = 1,260 ways

6 0
3 years ago
When the members of a family discussed where their annual reunion should take​ place, they found that out of all the family​ mem
Doss [256]

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

In which a are those that would only not go to a park, A \cap B are those who would not got to a park or to the beach, A \cap C are those who would not go to a park or to the famili cottage. And A \cap B \cap C are those that would not go to any of these places.

By the same logic, we have:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

This diagram has the following values:

a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)

The total number of family members is the sum of all these values:

T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family​ cottage.

This means that A \cap B \cap C = 5

1 would go to all three places. This means that d = 1.

8 would go to neither a park nor the family​ cottage

This means that:

A \cap C + (A \cap B \cap C) = 8

A \cap C = 3

8 would go to neither a beach nor the family​ cottage

B \cap C + (A \cap B \cap C) = 8

B \cap C = 3

7 would go to neither a park nor a​ beach

A \cap B + (A \cap B \cap C) = 7

A \cap B = 2

15 would not go to the family​ cottage

C = 15

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

15 = c + 3 + 3 + 5

c = 4

12 would not go to a​ beach

B = 12

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

12 = b + 3 + 2 + 5

b = 2

11 would not go to a​ park

A = 11

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

11 = a + 2 + 3 + 5

a = 1

Now, we can find the total number of family members.

T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)

T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5

T = 21

The total number of family members is 21.

5 0
2 years ago
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