A= 1
B= -5
C= 10
In order to figure it out, you can think about it like this..
The solution to the first expression - 7+3(9-4)^2÷5 is given as 22.
To get the answer correctly, one must follow rudimentary rules of operations which are coined into the acronym BODMAS.
<h3>What is BODMAS?</h3>
This is the order in which mathematical operations must be executed.
B = Bracket
O = Orders (that is Powers, Indices or roots)
D= Division
M = Multiplication
A = Addition
S = Subtraction
Now lets see how we got 22 from the first set of operations:
<h3>Operation 1 (Example)</h3>
7+3(9-4)^2÷5 =
7+3 (5)^2÷5=
7+3 * 25÷5 =
7+3*5=
7+15=
22
Following the BODMAS rule and the example in Operation 1 above, we can state the remaining answers as follows:
<h3>
Operation 2</h3>
12/3-4+7^2 = 49
<h3 /><h3>
Operation 3</h3>
(7-3)×3^3÷9 = 12
<h3>Operation 4</h3>
5(7-3)^2÷(6-4)^3-9 = 1
<h3>Operation 5</h3>
3×(7-5)^3÷(8÷4)^2-5 = 1
<h3>Operation 6</h3>
9+(3×10)/5×2-12 = 9
See the link below for more about Mathematical Operations:
brainly.com/question/14133018
Answer:
5083
Step-by-step explanation:
6^3 + 5(8 + 15)
Simplify the bracket first
6 x 6 x 6 + 5 (23)
216 + 5 (23)
221 (23)
221 x 23 = 5083
Answer:
1.We say a coin is fair if it has probability 1/2 of landing heads up and probability 1/2 of landing tails up. What is the probability that if we flip two fair coins, both will land heads up? It seems plausible that each should be equally likely. If so, each has probability of 1/4.
2.The probability of getting heads on the toss of a coin is 0.5. If we consider all possible outcomes of the toss of two coins as shown, there is only one outcome of the four in which both coins have come up heads, so the probability of getting heads on both coins is 0.25.
3.his states that the probability of the occurrence of two mutually exclusive events is the sum of their individual probabilities. As you can see from the picture, the probability of getting one head and one tail on the toss of two coins is 0
Step-by-step explanation:
About seven inches. Pretty sure
