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Len [333]
2 years ago
7

Please help me out on this​

Mathematics
1 answer:
belka [17]2 years ago
8 0
I have no idea hdhdhhdjdjd
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75% of students were boys. there 18 boys in the class wht is the total number of students
Kaylis [27]

Answer:

the answer is 24.

Step-by-step explanation:

5 0
2 years ago
A triangle has one leg that measures 12 cm and a hypotenuse 24 cm. Find the length of the other leg to the nearest tenth of cent
Minchanka [31]

9514 1404 393

Answer:

  20.8 cm

Step-by-step explanation:

The term "hypotenuse" suggests this triangle is a right triangle. Then the other leg can be found using the Pythagorean theorem:

  x^2 +12^2 = 24^2

  x^2 = 576 -144 = 432

  x = √432 ≈ 20.8 . . . cm

The other leg is about 20.8 cm.

_____

<em>Additional comment</em>

A right triangle with a short leg that is 1/2 the length of the hypotenuse is the "special" 30°-60°-90° right triangle. The longer leg is √3 times the length of the short leg: 12√3 ≈ 20.8 cm.

7 0
2 years ago
What terms can be combined with 4b? Select all that apply.
andriy [413]
16b and 3b are the answers
8 0
2 years ago
Read 2 more answers
There are two numbers, that sum up to 53. Three times the smaller number is equal to 19 more than the larger number. What are th
Taya2010 [7]

Answer:

x = 18  (Smaller number)

y = 35 (Larger number)

Step-by-step explanation:

To solve, begin by setting up a system of equations.

'x' being the smaller number, while

'y' is the larger number.

We can construct the equations:

x + y = 53

3x = y + 19

Rearrange the second equation to equal 'y':

3x - 19 = y

Plug this value of 'y' into the first equation:

x + 3x - 19 = 53

Combine like terms and simplify:

4x - 19 = 53

4x = 72

x = 18

Plug in this value for 'x' into an equation to solve for 'y':

18 + y = 53

y = 35.

7 0
3 years ago
Read 2 more answers
Which of the following expressions is equivalent to 5 x a? 5a. a^5. axaxaxaxa. 5^a
mixer [17]

Answer:

5a

Step-by-step explanation:

5 * a = 5a

Can you please mark this as brainliest ? thank you

6 0
3 years ago
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