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2 years ago

Solve for x. 3.3x = 29.04

Mathematics

2 answers:

disa [49]2 years ago

5 0

29.04 divide 3.3 =x
So therefore
x =8.8

Ilya [14]2 years ago

4 0

3.3x =29.04

to find x divide both sides by 3.3

x = 29.04 / 3.3

x = 8.8

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Answer:

The y-intercept in coordinate form is (0, 4.5) and represents the taxi pick-up fee. The equation is y=1.5x + 4.5.

Step-by-step explanation:

This question is asking for the slope and y-intercept of a linear equation. A linear equation makes a straight line based on a constant rate of change. For this problem, the cost per mile is the slope, while the independent variable 'x' is the number of miles and the dependent variable 'y' is the total cost. In order to first find slope, you need to use the two points given (7, 15) and (10, 19.5) to set up a change in y / change in x, or (19.5-15)/(10-7) or 4.5/3 which is 1.5. So the slope, or cost per mile is $1.50. To find the y-intercept (b), or the cost of the pick-up fee, simply fill in your equation y=1.5x + b with your other variables and solve for 'b'. So, 15 = (1.5 x7) + b. or 15 = 10.5 +b, subtract 10.5 from both sides of the equation to get b=4.5.

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deff fn [24]

Part (c)

We'll use this identity

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\$

to say

$\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\$

Similarly,

$\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

-------------------------

The key takeaways here are that

$\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

Therefore,

$2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\$

The identity is confirmed.

==========================================================

Part (d)

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\$

Similarly,

$\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\$

-----------------

We'll square each equation

$\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

and

$\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

--------------------

Let's compare the results we got.

$\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

Now if we add the terms straight down, we end up with $\sin^2(45+A)+\sin^2(45-A)$ on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how $\frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A)$ and similarly for the sin^2 terms as well.

The right hand side becomes $\cos^2(A)+\sin^2(A)$ but that's always equal to 1 (pythagorean trig identity)

This confirms that $\sin^2(45+A)+\sin^2(45-A) = 1$ is an identity

4 0

2 years ago