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3 years ago

How many different three digit locker combinations are possible if the digits are able to repeat?

1 answer:

8 0

1000, because if it’s number 0-9 you have 10 options for the first possible number, then multiply by 10 two more time for the second and third number .but if it’s wrong I’m srry rip ;P

You might be interested in

Solve the system of equations. Show ALL your work. Put your answer in the form of an ordered pair (x, y)

stiv31 [10]

This should be easy because we just have to use substitution method. Substitute the value of x from the second equation into the x of the first equation.
-4(2y) + 11y = 15
-8y + 11y = 15
3y = 15 ; y = 5
Substitute this value of y to either the first or second equation.
x = 2(5) = 10
The ordered pair is therefore (10,5).

6 0

3 years ago

What is the equation of a line with slope −23 and a y-intercept of −4 ?

Alona [7]

Answer:

its A

Step-by-step explanation:

the y intercept would be -4 and the slope would be -23, put it in y=mx+b form

8 0

3 years ago

What is the length of the missing leg? If necessary, round to the nearest tenth.

Marrrta [24]

Answer:

The length of the hypotenuse is 33 yards.

Step-by-step explanation:

We can solve for 'a' using Pythagoras theorem. Let's solve it.

65² = b² + 56²
=> 4225 = b² + 3136
=> b² = 1089
=> b = 33

Hence, the length of the hypotenuse is 33 yards.

Hoped this helped.

$BrainiacUser1357$

7 0

2 years ago

Pablo took out an 80/20 mortgage to buy a house that cost $140,000. The first (80%)

MArishka [77]

Answer:

$780.50

Step-by-step explanation:

I will update this if I figure how it's solved as well.

5 0

3 years ago

How do you solve 9| x+8 | + 10 < 55 & how is it graphed

mafiozo [28]

Hello!

First, let's write the problem.
$9\left|x+8\right|+10\ \textless \ 55|x+8|+10\ \textless \ 55$
Subtract 10 from both sides.
$9\left|x+8\right|+10-10\ \textless \ 55-10$
$9\left|x+8\right|\ \textless \ 45$
Divide both sides by 9.
$\frac{9\left|x+8\right|}{9}\ \textless \ \frac{45}{9}$
$\left|x+8\right|\ \textless \ 5$
When we got absolute value, we are going to get two results.
$x+8\ \textless \ 5$
and
$x+8\ \textgreater \ -5$

Let's solve the first one,
$x+8\ \textless \ 5x+8\ \textless \ 5$
Subtract 8 from both sides.
$x+8-8\ \textless \ 5-8$
$x\ \textless \ -3x\ \textless \ -3$

Let's solve the second one.
$x+8\ \textgreater \ -5$
Subtract 8 from both sides.
$x+8-8\ \textgreater \ -5-8$
$x\ \textgreater \ -13x\ \textgreater \ -13$

Let's combine the ranges.
$-13\ \textless \ x\ \textless \ -3$

The attachment shows how would you graph it.
You can feel free to let me know if you have any questions regarding this.
Thanks!

- TetraFish

3 0

3 years ago