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fiasKO [112]
2 years ago
14

If 4 wands are equivalent to 6 rands and 24 rands are equivalent to 8 fands, how many wands are equivalent to 5 fands?

Mathematics
1 answer:
nevsk [136]2 years ago
7 0

Answer:

22.5 Wands

Step-by-step explanation:

4 wands = 6 rands -> 6*4=24 so 4*4=16  -> 16 wands = 24 rands  (1 wand=1.5 rand)

24 rands = 8 fands -> 24/8 = 3 -> 3 rands = 1 fand

3 rands means 1.5*3=4.5 -> 4.5 wands = 1 fand -> 4.5*5=22.5 wands

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Chapter 12 F
marshall27 [118]

Answer:

(m+n) -a = r

Step-by-step explanation:

first, you have to find what m+n is because that is their total budget, then you have to remove how much their apartment costs from their budget, because r is the amount of free cash they have. once A is subtracted from m+n, you have r

8 0
3 years ago
What is (f + g)(x)?<br> f(x) = -2x^2 <br> g(x) = -2x^2 + 9x
andreyandreev [35.5K]

Answer:

-4x^2+9x

Step-by-step explanation:

Adding boh gives

(f+g)(x)=-2x^2+(-2x^2+9x)

=-2x^2-2x^2+9x

=-4x^2+9x

4 0
3 years ago
I can’t figure these three out I keep getting the wrong answers any help would be great....?
allochka39001 [22]
25.) D
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5 0
3 years ago
Read 2 more answers
The ratio of the angles in a triangle is 7:10:3 what is the measure of the smallest angle?
Sonbull [250]
The correct answer should be 3.
4 0
3 years ago
The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =
pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du

so that

v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du

\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is s(0)=0\,\mathrm m. Then its displacement over the time interval [0, 3] is

s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}

c) The total distance traveled is the integral of the absolute value of the velocity function:

s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt

v(t) for 0\le t and v(t)\ge0 for \cos^{-1}\left(-\frac29\right)\le t\le3, so we split the integral into two as

\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt

=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt

\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}

4 0
3 years ago
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