<h2>
Answer: -30y + 12u + 18 </h2>
<h3>
Step-by-step explanation:</h3>
<em> by multiplying each term in the parentheses by the term outside</em>
-6(5y-2u-3) = (-6 × 5y) + (-6 × - 2u) + (-6 × -3)
= -30y + 12u + 18
Answer:
Step-by-step explanation:
Joint variation problem solve using the equation y = kxz.
e ∝ fg
e=kfg
now substitute the values
The relationship will be:
Answer:
The correct options are;
1) Write tan(x + y) as sin(x + y) over cos(x + y)
2) Use the sum identity for sine to rewrite the numerator
3) Use the sum identity for cosine to rewrite the denominator
4) Divide both the numerator and denominator by cos(x)·cos(y)
5) Simplify fractions by dividing out common factors or using the tangent quotient identity
Step-by-step explanation:
Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;
tan(x + y) = sin(x + y)/(cos(x + y))
sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
Step-by-step explanation:
a. y = ¾x +6
b. y = 6/4 x -2
c. y = -2/4 x +3
d. y = -1/7 x -4
Px + qy = r
2px - qy = 2r
----------------add
3px = 3r
x = 3r/3p
x = r/p
