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3 years ago

Prime factor of 24 tree

Mathematics

1 answer:

ASHA 777 [7]3 years ago

5 0

$\begin{array}{c|cc}24&2\\12&2\\6&2\\3&3\\1\end{array}\\\\24=2\cdot2\cdot2\cdot3=2^3\cdot3$

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2(x−3)=(x−1)+7

Oksanka [162]

Answer:

x = 12

(getting to the 20 character limit, please ignore this)

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2 years ago

Read 2 more answers

Find the length of side AB in the right triangle shown.

Bogdan [553]

Side AB is 72 cm long.

Explanation:

To find the length of side AB you have to use this formula:

${a}^{2} + {b}^{2} = {c}^{2}$

*Remember there are two legs (a and b) and a hypotenuse (c) on a triangle. The longest side will be the hypotenuse and the other two will be the legs. For this triangle the hypotenuse is side BC while the legs are sides CA and side AB.

Plug in the numbers into the equation.

${65}^{2} + {b}^{2} = {97}^{2}$
For this triangle, you were given the hypotenuse (c) and one leg (a). To find the last length solve the equation and balance it so only b is left, like this:

$4225 + {b}^{2} = 9409$

$9409 - 4225 = 5184$

Soo...

${b}^{2} = 5184$

Then to get the side length, square root 5184.

$\sqrt{5184} = 72$

So the length of the last side is 72 cm.

5 0

3 years ago

A leaky faucet in kali's house drips water at a rate of 42 gallons per week. How many gallons does the faucet drip in three days

Ratling [72]

Amount of water that the faucet leaks in 1 week = 42 gallons

That means Amount of water that the faucet in Kali's house leaks in 7 days = 42 gallons So The amount of water that the faucet in Kali's house leaks in 1 day = (42/7) gallons = 6 gallons Then The amount of water that the faucet in Kali's house leaks in 3 days = 6 * 3 gallons = 18 gallons

So it can said from the above deduction that the amount of water leaked by the faucet in Kali's house is 18 gallons.

hope this help .-.

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3 years ago

Luke can build 2 tables in a day, and he has to build more than 18 wooden tables for a client. If he works for x days, what is t

VARVARA [1.3K]

2x > 18

2x/2 > 18/2

X > 9

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3 years ago

What are the number of integral solutions of the equation 2x+2y+z=20 such that x>=0 , y>=0 , z>=0?

zaharov [31]

$2x+2y+z=20\\
z=\dfrac{20}{2x+2y}\\
z=\dfrac{10}{x+y}$

Now, for z to be an integer, the sum x+y must be a divisor of 10.
It has to be a positive divisor since z≥0. Also x≠y.

$x+y=1 \\
x=1-y\\
$
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.

$x+y=2\\
x=2-y\\$
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.

$x+y=5\\
x=5-y$
y can be 0,1,2,3,4 or 5 - 6 solutions

$x+y=10\\
x=10-y$
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.

2+2+6+10=20 solutions in total.

6 0

3 years ago