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2 years ago

How could you solve this problem?

Mathematics

2 answers:

Kamila [148]2 years ago

7 0

2x3x4 = 24 so the only statements that are correct are the 3rd and 4th ones as they are the only ones that equal 24.

Lubov Fominskaja [6]2 years ago

4 0

Umm try 5xtxxtxtxtxtxtxtxxttxxtxtxtxtxtxttxtxtxxtxtxtxtxtxt txtxtxxt txtxtxtxtxt txtxtxtx txtxtxtxt cat txt

You might be interested in

What percentage is $12.10 out of $58.60

jasenka [17]

Answer:

12.10 out of 58.60 is 20.648464163822524 or 21%

Step-by-step explanation:

work out the difference (increase) between the two numbers you are comparing

Increase = New Number - Original Number

Divide the increase by the original number and then multiply the answer by 100

% increase = Increase ÷ Original Number × 100.

5 0

3 years ago

Please help me ASAP yall!!!

ASHA 777 [7]

Answer:

Step-by-step explanation:

<C+<D = 180 since they form a straight line

5x+20 +3x = 180

8x+20 = 180

Subtract 20 from each side

8x+20-20 = 180-20

8x = 160

Divide by 8

8x/8 = 160/8

x = 20

We want to find angle D

<D = 3x= 3*20 = 60

8 0

3 years ago

What is an inequality for x <12

natka813 [3]

X < 12 its actually very simple , just see that the x is on one side and nothing can be moved

4 0

4 years ago

Write the missing decimal so that each pair adds to 1. 2/5+decimal =1

I am Lyosha [343]

Answer:

0.6

Step-by-step explanation:

We can get the answer by subtracting the decimal by one.

2/5 = 0.4

1 - 0.4 = 0.6

4 0

2 years ago

At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the firs

rewona [7]

Answer:

4032 different tickets are possible.

Step-by-step explanation:

Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.

To find : How many different tickets are possible ?

Solution :

In the first race there are 9 ways to pick the winner for first and second place.

Number of ways for first place - $^9C_1=9$

Number of ways for second place - $^8C_1=8$

In the second race there are 8 ways to pick the winner for first and second place.

Number of ways for first place - $^8C_1=8$

Number of ways for second place - $^7C_1=7$

Total number of different tickets are possible is

$n=9\times 8\times 8\times 7$

$n=4032$

Therefore, 4032 different tickets are possible.

8 0

3 years ago