The question is saying, how many three digit numbers can be made from the digits 3, 4, 6, and 7 but there can't be two of the same digit in them. For example 346 fits the requirements, but 776 doesn't, because it has two 7s.

Okay, on to the problem:

We can do one digit at a time.

First digit:

There are 4 digits that we can choose from. (3, 4, 6, and 7)

Second digit:

No matter which digit we chose for the first digit, there is only going to be 3 of them left, because we already chose one, and you can't repeat that same digit. So there are 3 options.

Third digit:

Using the same logic, there are only 2 options left.

We have 4 choices for the first digit, 3 choices for the second, and 2 for the third.

Hence, this is 4 * 3 * 2 = 24 three-digit numbers that can be made.

8 0

2 years ago

ΔHAT is similar to ΔCAN.

LenKa [72]

In similar triangles, the ratios of corresponding sides are equal.

This means

y/6=5/(4+5)

Cross multiply

y=5*6/(4+5)=30/9=10/3 (or 3 1/3)

6 0

3 years ago

Read 2 more answers

Find the quotient of 5+4i/6+8i , and express it in the simplest form

Sergio [31]

Remember: i² = -1

$\frac{5+4i}{6+8i}= \frac{(5+4i)(6-8i)}{(6+8i)(6-8i)}= \frac{30-40i+24i-32i^2}{36-64i^2}= \frac{30-16i-32(-1)}{36-64(-1)}= \\ \\ = \frac{30-16i+32}{36+64}= \frac{62-16i}{100}=0.62-0.16i$

Hope this helps

7 0

3 years ago

Expanding log functions<br> ㏒3 (3(x+1)(x+2))

Oksanka [162]

Log(ab)=log(a)+log(b)
log_a(a)=1
so

$log_3(3(x+1)(x+2))=log_3(3)+log_3(x+1)+log(x+2)=$ $1+log_3(x+1)+log_3(x+2)$

6 0

3 years ago