Answer:
x = 5/13
Step-by-step explanation:

<span>The <u>correct answers</u> are:
A ray is a bisector of an angle if and only if it splits the angle into two angles; and
A) I can afford to buy a ticket.
Explanation<span>:
For the first question, the first three answers are very specific and true:
A whole number is odd if it is not divisible by 2, and a number is not divisible by 2 if it is odd;
an angle is straight if its measure is 180 degrees, and the measure of an angle is 180 degrees if it is a straight angle;
a whole number is even if it is divisible by 2, and a number is divisible by 2 if it is even.
However, with the fourth choice, we are missing a key word in the definition. A ray is a bisector of an angle if and only if it splits the angle into two <u>CONGRUENT</u> angles. It is not just a ray that cuts an angle into two pieces, the pieces must be equal.
For the second question, the Law of Detachment says if our conditional "if p, then q" is true and p is true, then q must also be true.
For this question, "I can go to the concert if I can afford to buy a ticket" is true as well as "I can go to the concert." This means "I can afford to buy a ticket" must be true as well.</span></span>
Solution to the question:
Number of blocks stacked: 30
Height of tower: 64 inches
Height of each block: ?
To find the height of each block, we divide the total height of the tower with the number of blocks stacked
Therefore,
Height of each block: 64/30 = 2.13 inches
So, the height of each block is 2.13 inches.
Answer:
20$ is the answer of your question
Step-by-step explanation:
Let n be the number of nickels that Casey has.
We are told that Megan has 4 times as many nickels as Casey has, therefore, the number of nickels that Megan has will be 4*n.
Now we will add the amounts of nickels that Casey and Megan have to find the total amount of nickels.

Therefore, the expression
will represent the total number of nickels that Casey and Megan have .
Upon combining like terms we will get,


Therefore, Casey and Megan have 5n nickels.