How do I do these problems?

Mathematics

1 answer:

Novosadov [1.4K]2 years ago

7 0

A midsegment is half as long as the one it's parallel to.

2. x = 16/2 = 8

3. y+4 = 6
y = 2

4. z/2 = 15/2
z = 15

5. (5/2)x + 4 = (6x +4)/2
2 = x/2
x = 4 . . . . . . matches selection b.

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The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this prope

sineoko [7]

Answer:

$\large \boxed{17}$

Step-by-step explanation:

Let's call the numbers "tu", where t is the tens digit and u is the units digit.

We can solve this by the "brute force" method — examining each number to see if it satisfies the conditions.

$\begin{array}{cccl}\mathbf{u}&\mathbf{tu}& \textbf{Divisible by u} & \\0 &20, 30, 40 & 0 & \text{Division by zero is impossible}\\1 & 11, 21, 31, 41 & 4& \text{All numbers are divisible by one}\\2 & 12, 22, 32, 42 & 4 & \text{All even numbers are divisible by two}\\3 & 13, 23, 33, 43 & 1 &\text{Only 33 is divisible by three}\\4 & 14, 24, 34, 44 & 2 & \text{Only 24 and 44 are divisible by four}\\5 & 15, 25, 35, 45 & 4 & \text{Numbers ending in 5 are divisible by five}\\\end{array}$

$\begin{array}{cccl}6 & 16, 26, 36, 46 & 1 & \text{Only 36 is divisible by six}\\7 & 17, 27, 37, 47 & 0 & \text{None of the numbers is divisible by seven}\\8 & 18, 28, 38, 48 & 1 & \text{Only 48 is divisible by eight}\\9 & 19, 29, 39, 49 & 0 & \text{None of the numbers is divisible by nine}\\& \textbf{TOTAL =} &\mathbf{17}& \\\end{array}\\\text{$\large \boxed{\mathbf{17}}$ numbers between 10 and 50 are divisible by their units digit}$