What is the question? It doesn’t show
Answer:
cosM ≈ 0.3048
Step-by-step explanation:
remember SOH CAH TOA = sine: opposite/hypotenuse; cosine: adjacent/hypotenuse; tangent: opposite/adjacent
cos = CAH
cosM= a / c
cosM = 4.0 / 13.1244
cosM = 0.30477
cosM ≈ 0.3048
Answer:
x = - 56/9 or -6.2
Step-by-step explanation:
Answer:
Amira used 4 balloons for each balloon animal.
Step-by-step explanation:
From the table, we can conclude that:
For 20 animals being sold by her, the leftover balloons were 180.
Then there is an increase in the number of animals being sold. She now sells 29 animals. So, the increase in animals being sold is given as:
29 - 20 = 9.
So, 9 more animals were sold. The balloons used for these 9 animals can be obtained by taking the difference of the leftover balloons for the two days.
The difference of the leftover balloons = 180 - 144 = 36.
So, 36 balloons were used for selling 9 more animals.
Similarly, there is an increase of another 9 animals when 38 animals were sold as 38 - 29 = 9. Also, the balloons used in this case is 36 only as the difference of the leftover balloons is 36.
144 - 108 = 36
Therefore, we conclude that for selling 9 balloon animals, Amira used 36 balloons.
Number of balloons used for 9 balloon animals = 36
∴ Number of balloons used for 1 balloon animal =
(Unitary method)
Thus, Amira used 4 balloons for each balloon animal.
The two numbers are 12 and 15
<em><u>Solution:</u></em>
Let the smaller number be "x"
Let the larger number be "y"
Given that,
<em><u>The sum of two numbers is 27</u></em>
Therefore, we frame a equation as:
Smaller number + Larger number = 27
x + y = 27 ------ eqn 1
<em><u>The larger number is 3 more than the smaller number</u></em>
Larger number = 3 + smaller number
y = 3 + x ------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
<em><u>Substitute eqn 2 in eqn 1</u></em>
x + 3 + x = 27
2x + 3 = 27
2x = 27 - 3
2x = 24
<h3>x = 12</h3>
<em><u>Substitute x = 12 in eqn 2</u></em>
y = 3 + 12
<h3>y = 15</h3>
Thus the two numbers are 12 and 15