Answer:
10.9
Step-by-step explanation:
Formula = Number x 100
Percent = 6 x 100
55 = 10.91
Following shows the steps on how to derive this formula
Step 1: If 55% of a number is 6, then what is 100% of that number? Setup the equation.
6
55% = Y
100%
Step 2: Solve for Y
Using cross multiplication of two fractions, we get
55Y = 6 x 100
55Y = 600
Y = 600
100 = 10.91
Answer:
B and E
Step-by-step explanation:
To verify which are solutions, substitute the given values into the left side of the equation and if equal to the right side then they are solutions.
A
2(20)² + 18(20) = 800 + 360 = 1160 ≠ 20 ← not a solution
B
2(1)² + 18(1) = 2 + 18 = 20 ← solution
C
2(- 2)² + 18(- 2) = 2(4) - 36 = 8 - 36 = - 28 ≠ 20 ← not a solution
D
Is a repeat of C
E
2(- 10)² + 18(- 10) = 2(100) - 180 = 200 - 180 = 20 ← solution
F
2(- 1)² + 18(- 1) = 2(1) - 18 = 2 - 18 = - 16 ≈ 20 ← not a solution
Then B and E are solutions to the equation
Answer:
93
Step-by-step explanation:
Key :
A1 = Algebra 1
A2 = Algebra 2
Alright so basically lets first look at the info they gave us :
We have 5 more than twice as many students taking A1 than we do A2.
We have 44 students taking A2.
And we need to find the least amount of students that could be taking A1.
So we need to take the amount of students taking A2 (44) and double it to find the amount taking A1.
So we can do 44 x 2 = 88 to get this.
But the problem also states there is 5 more then twice the number of students taking A2.
So we have that 88 but now we just need to add 5 to make up for them telling us that in the problem.
So :
88 + 5 = 93
Our final answer and least amount of students taking A1 is 93 students.
Answer:
15tan 30
Step-by-step explanation:15 tan30°
Remember SOHCAHTOA.
tanθ =
opposite
adjacent
tan30° =
x
15
x = 15 tan30°
Answer:
45
Step-by-step explanation:
Simply start by listing out the numbers from smallest to largest.
We start out with the smallest number following this information being the number 10. The next are 21 and 20. The next are 32, 31, and 30. We can start to see a pattern. If the ten's digit is n, then there are n numbers within that ten-group that have a greater 10's digit than unit's digit. The largest ten-group (90 to 98) has 9 integers following the rules established. Thus, the number of 2-digit integers that have ten's digits greater than units digit is 1+2+3+...+9=45
