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3 years ago

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).

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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

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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

$(A \cup B) + C = 1$