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Viktor [21]
3 years ago
15

Segment A'B' is parallel to segment AB.What is the length of segment AB?

Mathematics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

7.5

Step-by-step explanation:

5/6=x/9

cross multiply

6x=45

x=45/6

x=7 3/6 = 7 1/2 = 7.5

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Express the product of 2x^2+7x-10 and x+5 in standard form
zimovet [89]

Answer:

2x³+17x²+25x-50

Step-by-step explanation:

Standard Form: All your numbers and variables in order of the highest exponent.

(2x²+7x-10)(x+5)

We want to distribute x and 5 to every single value.

2x³+7x²-10x+10x²+35x-50

Simplify.

2x³+17x²+25x-50

This is standard form because the highest values are on the left (x³ is larger than x²).

4 0
2 years ago
4 tomatoes cost 3.40. What is the unit rate?
Romashka-Z-Leto [24]
3.40/4= $0.85 per tomato
8 0
3 years ago
Read 2 more answers
PLEASE HELP !!!!!!! WILL MARK BRAINLIEST FOR A REAL ANSWER
rjkz [21]

Answer:

A. Find the maximum height of the acrobat.

Step-by-step explanation:

3 0
3 years ago
Solve system of equation using elimination by addition.<br><br> Will give brainliest!!
DedPeter [7]

Answer:

x = 2, y = -3

Step-by-step explanation:

Add the 2 equations together, then solve for x.

(2x + 2y) + (3x - 2y) = (-2) + (12)\\ 2x + 3x + 2y - 2y = -2 + 12\\5x = 10\\x = 2

Now that we know the value of x, we can easily find y by substitution.

3x - 2y = 12\\3(2) - 2y = 12\\6 - 2y = 12\\-2y = 6\\y = -3

These are the answers

- Kan Academy Advance

5 0
2 years ago
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
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