Based on the percentage that passed English and those who passed Mathematics and those who failed and passed both, the total number of students who appeared in the examination are 60 students.
The number of students who passed only in Math are 12 students.
<h3>What number of students sat in the exam?</h3>
This can be found as:
= Total who passed English only + Total who passed Math only + Total who failed both + Total who passed both
Assuming the total is n, the equation becomes:
n = 0.75n - 21 + 0.55n - 21 + 21 + 0.05n
n = 1.35n - 21
21 = 0.35n
n = 21 / 0.35
= 60 students
The number who passed mathematics only is:
= (60 x 55%) - students who passed both
= 33 - 21
= 12 students
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Original coordinates of the points:
A (8,15) ; B (12,13) ; C (8,10)
Dilated scale factor of 3.
A ⇒ 3x = 3(8) = 24 ; 3y = 3(15) = 45 ⇒ A' (24,45)
B ⇒ 3x = 3(12) = 36 ; 3y = 3(13) = 39 ⇒ B' (36, 39)
C ⇒ 3x = 3(8) = 24 ; 3y = 3(10) = 30 ⇒ C' (24, 30)
The given image forms a right triangle. So, I'll get the short leg and long leg of the right triangle to solve for the hypotenuse, length of CB.
Short leg: y value of B and C
39 - 30 = 9
Long leg: x value of B and C
36 - 24 = 12
a² + b² = c²
9² + 12² = c²
81 + 144 = c²
225 = c²
√225 = √c²
15 = c
The length of CB is 15 units.
Answer:
Step-by-step explanation:
The opposite side (the one not connected to A) = 4
The hypotenuse is 5
The adjacent side needs to be found for the cosine and the tangent.
a^2 + b^2 = c^2
a = opposite side = 4
b = adjacent side = ?
c = hypotenuse = 5
4^2 + x^2 = 5^2
16 + x^2 = 25
x^2 = 25 - 16
x^2 = 9
x = sqrt(9)
x = 3
cos(A) = adjacent / hypotenuse = 3/5
Tan(A) = opposite / adjacent = 4/3
cos(A) + tan(A) = 3/5 + 4/3
cos(A) + tan(A) = 9/15 + 20/15 = 29/15
