Answer:
The trigonometric equation (sin Θ − cos Θ)^2 − (sin Θ + cos Θ)^3 can be simplified by:Using x for Θ: (sinx - cosx)^2 - (sinx + cosx)^2 = (sin^2 x - 2sinxcosx + cos^2 x) - (sin^2 x + 2sinxcosx + cos^2 x) = - 2 sinx cosx - 2 sinx cosx = - 4 sinx cosx = - 2sin(2x)
Step-by-step explanation:
Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:
<em>y</em> = -1 - <em>x</em> ==> <em>x</em> = -1 - <em>y</em>
<em>y</em> = <em>x</em> - 1 ==> <em>x</em> = 1 + <em>y</em>
So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.
This means the area is given by the integral,

The integral with respect to <em>x</em> is trivial:

For the remaining integral, integrate term-by-term to get

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.
9. 130 = x + 137
answer x = -7
10. 9x-3 = 4 + 8x
9x = 7 + 8x
answer x = 7
Answer:
Step-by-step explanation:
10
Answer:
The algebraic expression is v = 2x
v is the volume of the sugar syrup and
x is the mass of sugar in grams.
Step-by-step explanation:
Let x be the mass of sugar in grams and v be the volume of sugar syrup.
So, mass of sugar in grams/volume of sugar syrup × 100 % = 50 %
x/v × 100 % = 50 %
x/v = 50/100
x/v = 1/2
v = 2x
So, the algebraic expression required is v = 2x where v is the volume of the sugar syrup and x is the mass of sugar in grams.