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3 years ago

Can u combine 3xy with 3y

Mathematics

2 answers:

Wewaii [24]3 years ago

3 0

Can the answer is 6 xy2

Triss [41]3 years ago

3 0

No because the bases aren't the same 3xy doesn't have the same base as 3y. The x is not shown in both set of givens.

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Can someone help me please

Novay_Z [31]

Answer:

B. x = -5 +/- √30.

Step-by-step explanation:

2x^2 + 20x = 10

2(x^2 + 10x) = 10

2[ (x + 5)^2 - 25 ] =10

2(x + 5)^2 - 50 = 10

2(x + 5)^2 = 60

(x + 5)^2 = 30

x + 5 = +/- √30

x = -5 +/- √30.

6 0

3 years ago

Ms. Moneyhun's son, Sam, just learned to crawl. He can crawl 20 ft in one minute. How many feet can Sam crawl in one hour? (Hint

almond37 [142]

Answer:

1200 ft in one hour

Step-by-step explanation:

The ratio of Sam crawling 20 ft in one minute can be written as:

20:1

Multiply the ratio by 60 to get the minutes to 60 minutes, because there are 60 minutes in an hour:

20 : 1

×60 ×60

To get:

1200:60 ⇒ This means Sam can crawl 1200 ft in 60 minutes, which is the same as 1200 ft in one hour.

<em>Hope this helps!</em> :)

6 0

3 years ago

Need help right now!!!!!!

Reptile [31]

Answer:

44.75

I believe

8 0

3 years ago

Read 2 more answers

A fast-food company is interested in knowing the probability of whether a customer viewed an advertisement for their new special

lord [1]

Answer:

45% probability that a randomly selected customer saw the advertisement on the internet or on television

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 customer saw the advertisement on the internet.

B is the probability that a customer saw the advertisement on television.

We have that:

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

In which a is the probability that a customer saw the advertisement on the internet but not on television and $A \cap B$ is the probability that the customers saw the advertisement in both the internet and on television.

By the same logic, we have that:

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

12% saw it on both the internet and on television.

This means that $A \cap B = 0.12$

20% saw it on television

This means that $B = 0.2$

37% of customers saw the advertisement on the internet

This means that $A = 0.37$

What is the probability that a randomly selected customer saw the advertisement on the internet or on television

$A \cup B = A + B - (A \cap B) = 0.37 + 0.2 - 0.12 = 0.45$

45% probability that a randomly selected customer saw the advertisement on the internet or on television

6 0

2 years ago

Central High wants to estimate the number of seniors who plan to go to a 4-year college. Answer the following.

Elodia [21]

Answer:

390 seniors

Step-by-step explanation:

13/20=.65

600x .65=390

4 0

2 years ago

Read 2 more answers