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vichka [17]
3 years ago
5

Solve for f. 8 = 2f + 2

Mathematics
1 answer:
tiny-mole [99]3 years ago
5 0

Answer:

f = 3

Step-by-step explanation:

8 = 2f + 2

-2         -2

6 = 2f

/2    /2

3 = f

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How far is a football kicked if its y=-0.03x^2+1.53x
weqwewe [10]
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3 years ago
(06.04 MC)
Andru [333]

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

◉ \large\bm{ -4}

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

Before performing any calculation it's good to recall a few properties of integrals:

\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx +  \int_{a}^bmdx}

\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a  f(x)dx= \int^c _a f(x)dx+  \int^{b} _c  f(x)dx }

So we apply the first property in the first expression given by the question:

\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}

And we solve the second integral:

\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }

\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx  = 2 \int ^3_{-2} f(x)dx +   2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }

Then we take the last equation and we subtract 10 from both sides:

\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}

\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx  = 8}

And we divide both sides by 2:

\small\longrightarrow \sf{\dfrac{2  {  \int}^{3} _{2}  }{2}  =  \dfrac{8}{2} }

\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}

Then we apply the second property to this integral:

\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}

Then we use the other equality in the question and we get:

\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx  =  2 \int ^3_{-2} f(x)dx  = 8 +  2 \int ^3_{-2} f(x)dx  = 4}

\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}

We substract 8 from both sides:

\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}

• \small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}

7 0
2 years ago
Can anyone explain those marked steps in case (iii)
Tema [17]
\cos\dfrac{3x}2=\cos\left(\dfrac\pi2+\dfrac x2\right)
\implies \dfrac{3x}2=2n\pi+\dfrac\pi2+\dfrac x2

follows from the fact that the cosine function is 2\pi-periodic, which means \cos x=\cos(2\pi+x). Roughly speaking, this is the same as saying that a point on a circle is the same as the point you get by completing a full revolution around the circle (i.e. add 2\pi to the original point's angle with respect to the horizontal axis).

If you make another complete revolution (so we're effectively adding 4\pi) we get the same result: \cos x=\cos(4\pi+x). This is true for any number of complete revolutions, so that this pattern holds for any even multiple of \pi added to the argument. Therefore \cos x=\cos(2n\pi+x) for any integer n.

Next, because \cos(-x)=\cos x, it follows that \cos x=\cos(2n\pi-x) is also true for any integer n. So we have

\implies \dfrac{3x}2=2n\pi\pm\left(\dfrac\pi2+\dfrac x2\right)

The rest follows from considering either case and solving for x.
5 0
3 years ago
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