Answer:
63 monocotyledons
Step-by-step explanation:
some abbreviations ill be using:
monocotyledons=m
dicotyledons=d
3m / 4d
x / 84d
you need to figure out what x is
you can divide 84/4 and get 21
now that you have 21, you can multiple that number by 3
21x3=63
therefore, there are 63 monocotyledons
Answer:
a = 22
b = 31
c = 13
Step-by-step explanation:
The sum is the same in each row, column, and diagonal.
One of the diagonals is already complete. The sum is:
16 + 25 + 34 = 75
So the first row adds up to 75:
a + 37 + 16 = 75
a = 22
The second row adds up to 75:
19 + 25 + b = 75
b = 31
And the third row adds up to 75:
34 + c + 28 = 75
c = 13
We can check our answer by finding the sum of each column and the other diagonal.
22 + 19 + 34 = 75
37 + 25 + 13 = 75
16 + 31 + 28 = 75
22 + 25 + 28 = 75
The missing reason is (d) Add the fractions together on the right side of the equation
<h3>How to complete the missing reason?</h3>
From the statements, we have the following equation:
x^2 + b/a x + (b/2a)^2 = -4ac/4a^2 + b^2/4a^2
Next, we add the fractions on the right-hand side of the equation.
This gives
x^2 + b/a x + (b/2a)^2 = [-4ac + b^2]/4a^2
The above means that the last statement is gotten by adding the fractions on the right-hand side of the equation.
Hence, the missing reason is (d) Add the fractions together on the right side of the equation
Read more about quadratic equations at:
brainly.com/question/1214333
#SPJ1
Answer: The linear system of equations that models this scenario are
T + B = 15
3T + 2B = 38
Step-by-step explanation:
Let T represent the number of tricycles.
Let B represents the number of bicycles ordered.
The recreation center ordered a total of 15 tricycles and bicycles from a sporting good store. It is expressed as
T + B = 15
A tricycle has three wheels. A bicycle has 2 wheels. The number of wheels for all tricycles and bicycles total 38. The expression would be
3T + 2B = 38
Answer:
well yes, but no
Step-by-step explanation:
we can post other things than math but we shouldn't post it in the math forum
