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3 years ago

Can anyone help me with question number 3

Mathematics

1 answer:

ira [324]3 years ago

7 0

The answer is: There are 595 calories in a banana split

To find this we first need to set up our equation:

$80= \frac{1}{7}x-5$ , where x= calories in a banana split

$85= \frac{1}{7}x$ ----------------Add 5 to both sides

595=x-------------------Multiply 7 to both sides to get rid of the fraction

So, there are 595 calories in a banana split

Hope this helps! :)

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3 years ago

Read 2 more answers

Prove it please answer only if you know

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.

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