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Phoenix [80]
3 years ago
8

Combine like terms. -5y - 5x + 5y + 5y - x + 2 - 2

Mathematics
1 answer:
Karolina [17]3 years ago
6 0

Answer:

-6x + 5y

Explanation:

-5y - 5x + 5y + 5y - x + 2 - 2

-5x - x + 10y - 5y + 2 - 2

-6x + 5y + 0

-6x + 5y

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The equation of a line given two points needs to be found. Samuel claims that slope-intercept form will generate the equation an
Nikitich [7]

Answer:

Helena is correct.

Step-by-step explanation:

The slope intersept form of the line is

y = m x + c

where, m is the slope and c is the intersept.

The point slope form of the equation is

y - y'= \frac{y''-y'}{x''-x'}\times (x-x')

So, if there are two points are given, the equation of line is found by the point slope form.

So, Helena is correct.

8 0
3 years ago
A yard is approximately 5.68 x 10-4 miles. How is this number expressed in standard decimal notation?
belka [17]

Answer:

C: 0.000568 miles

Step-by-step explanation:

Essentially, the exponent on the 10 is how many places to move the decimal point, and negative versus positive dictates whether to move it right or left respectively.

4 0
3 years ago
Read 2 more answers
What percentage of 440 is 325
GuDViN [60]

The answer would be 73.86

3 0
2 years ago
Read 2 more answers
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0
3 years ago
Z divided by 3-2z + 4(z-8) when z = 18 <br><br> A. -122<br> B. 6<br> C. -66<br> D. 10
murzikaleks [220]

Everywhere you see z, replace the letter with 18.

z/[(3- 2z)] + 4(z - 8)

18/[(3 - 2(18))] + 4(18 - 8)

You finish.

3 0
3 years ago
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