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snow_tiger [21]
3 years ago
7

Maya made $154 for 11 hours. At the same rate, how many hours would she have to work to make $98

Mathematics
1 answer:
Artist 52 [7]3 years ago
3 0

it would be 7 hours.



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Urgent
kirza4 [7]

Answer:

Step-by-step explanation:

F(x) = 2^x

64 = 2^x

2^6 = 2^x

x = 6

f(x) = 2^x

log (f(x)) = x log 2

x = log (f(x)) / log 2 = log_2 (f(x))

Therefore, the required logarithmic function is x = log_2 (f(x))

Click to let others know, how helpful is it

8 0
3 years ago
Read 2 more answers
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
A 2-digit number with x units and 7 less units than tens.
Hatshy [7]
Each two digit number has two numbers (duh!). Let's allow the tens digit to be x and the units digit to be y. Tens digit is 3 less than the units digit: x = y-3 Original number is 6 more than 4 times the sum of the digits: 10x+y-6 = 4x + 4y This gives us simultaneous equations!First let's clear the mess: 1. x= y-32. 6x-3y=6 Substitute 1 into 2: 6(y-3) -3y =66y - 18 - 3y = 6
 3y = 24y = 8 Our units digit is 8 Substitute y= 8 into 1. x = y - 3x = 5 Our tens digit is 5 Therefore, our number is 58
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3 years ago
How many square inches does an "11x13" area encompass
Juliette [100K]
143 square inches         <span> </span>
3 0
4 years ago
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Would that he the answer or option A ?
sammy [17]
Your answer is correct
4 0
3 years ago
Read 2 more answers
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