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BaLLatris [955]
3 years ago
9

If f(x) = 6(9)^x then f(1/2) =?

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
F(x) = 6(9)^x

f(1/2) = 6(9)^(1/2)

f(1/2) = 6(9)^(0.5)

f(1/2) = 6*(9^0.5)               9^0.5 = √9 = +3 or -3.

f(1/2) = 6*3 = 18          or       6*-3 = -18

f(1/2) = 18   or   -18     

Hope this helps.
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kenny6666 [7]

The inequality is given by 20x + 25y ≤ 1100 and x + y > 50

<h3>What is an inequality?</h3>

An inequality is an expression that shows the non equal comparison of two or more numbers and variables.

Let y represent the number of soccer balls and x represent the number of volleyballs.

The coach can spend a maximum of $1,100, hence:

20x + 25y ≤ 1100

Also, The coach plans to order at least 50, hence:

x + y > 50

The inequality is given by 20x + 25y ≤ 1100 and x + y > 50

Find out more on inequality at: brainly.com/question/24372553

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John does 5 pushups on the first day of a 30-day month, and then increases the number of pushups by 2 pushups a day. how many pu
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He has done 70 push ups because by the 15th he has done 35 push ups add 15 more days and u get 70 .
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A) 2(x+3)=x-4 <br>Could anyone help?​
oee [108]
If solving for x you will first need to combine like terms. Then after getting the amount of numbers and amount of x numbers you will divide the x number from both. After doing that you would get X=-10
7 0
3 years ago
the length of a rectangle is 2 cm less than five times the width. the area of the rectangle is 16 cm^2. find the length and widt
ahrayia [7]
<span>length = L
</span>width = w

L = (5w-2) cm

w(5w-2)=16 \\ 5w^2-2w-16=0 \\ D=b^2-4ac=(-2)^2-4*5*(-16)=324 \\ w_{1,2}= \frac{-bб \sqrt{D} }{2a}  \\ w_1= \frac{2- \sqrt{324} }{2*5}= \frac{2-18}{10}=-1.6 \ \ \O \\ w_2=  \frac{2+18}{10}=2 cm

width = 2 cm
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8 0
3 years ago
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
3 years ago
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