Ask question

Ask question

All categories

2 years ago

14

Shaul made a password that consists of one letter followed by two digits. The two digits are different. How many possible passwo

rds did Shaul choose from?

1 answer:

Novay_Z [31]2 years ago

6 0

278. I got It By Counting The Letters On Meh Keyboard As Many Times As The #'s On My Keyboard.

You might be interested in

Enter what that means? <br> Help someone.

Luda [366]

$\frac{x^2}{3}$ is the same as $\frac{1}{3}x^2$

Dividing by 3 is the same as multiplying by the fraction 1/3

7 0

3 years ago

Pleaseeeeee helpppppppppplpoo

nataly862011 [7]

The answer is c: y = 10.29(0.7)^x.

4 0

3 years ago

X^2 -9X = 0. Factor and use the Zero property to solve

Veseljchak [2.6K]

The first step of factoring is to try to factor out a common factor.
The terms x^2 and -9x have the factor x in common.
Factor out x from both terms.

x^2 - 9x = 0

x(x - 9) = 0

Now you have a product of fully factored terms equaling zero, so you can apply the zero product property to solve.

x = 0 or x - 9 = 0

x = 0 or x = 9

Answer: x = 0 or x = 9

5 0

2 years ago

Are the following triangles similar? Justify your answer.

alukav5142 [94]

Yes. The triangles are both right triangles, which means they both have a single right angle.

3 0

3 years ago

Find the error with this proof and explain how it mat be corrected in order to clearly prove the equation.

enyata [817]

Answer:

Step-by-step explanation:

It is true that for any given odd integer, square of that integer will also be odd.

i.e if $m$ is and odd integer then $m^{2}$ is also odd.

In the given proof the expansion for $(2k + 1)^{2}$ is incorrect.

By definition we know,

$(a+b)^{2} = a^{2} + b^{2} + 2ab$

∴ $(2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k$

Now, we know $4k^{2}$ and $4k$ will be even values

∴ $4k^{2} + 1 + 4k$ will be odd

hence $(2k + 1)^{2}$ will be odd, which means $m^{2}$ will be odd.

7 0

2 years ago