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Marrrta [24]
3 years ago
6

6n^2-30n+42 can someone help me

Mathematics
1 answer:
sesenic [268]3 years ago
7 0

Answer:

\frac{5+i\sqrt{3}}{2}}

Step-by-step explanation:

I assume you want the roots. Here we go:

Use the quadratic formula: \frac{-(-30) plus or minus \sqrt{(-30)^2-4(6)(42)} }{2\cdot6}.

The discriminant is 900 - 1008, or -108 (so we will need i, the complex number that means \sqrt{-1}).


This means on the top you get 30 + 6isqrt3 and on the bottom you get 12.

So the fraction is \frac{30+6i\sqrt{3}}{12}=\boxed{\frac{5+i\sqrt{3}}{2}}.\blacksquare

Note: This is just one of the roots, I am sure you can find the second one!

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What is the area of the yellow section?
nignag [31]

Answer:

98m^{2}   i hope this helps!   :)

Step-by-step explanation:

given   7 * 14 = ?

multiply 14 by 7   14 * 7 = 98

98m^{2}

8 0
3 years ago
Read 2 more answers
Gelppppppppp meeeeeeee
Alja [10]

Answer:

3/5

Step-by-step explanation:

sin is opposite over hypotenuse

find hypotenuse first -> Hypotenuse = 100 m

(notice that it is a 3-4-5 triangle)

60/100 = 6/10 = 3/5

3 0
3 years ago
At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount. E
spayn [35]

At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount.

Let x be  the amount of one of Derek's withdrawals

Each of John's withdrawals was $5 more than each withdrawal that Derek made.

x + 5 is t the amount of one of John's withdrawals

Derek made 7 withdrawals

So amount withdraw 7 times = 7x

John made 5 withdrawals

So amount withdraw 5 times = 5(x+5)

Both Derek and John withdrew the same amount of money in the end

(A) 7x = 5(x+5)

(B) Solve for x

7x = 5x + 25

Subtract 5x from both sides

2x = 25

Divide by 2

x = 12.5

(C) check your solution

we plug in 12.5 for x in 7x= 5x + 25

7(12.5) = 5(12.5) + 25

87.5 = 62.5+ 25

87.5 = 87.5

(D) Each brother withdrawal 87.5 dollars



6 0
2 years ago
Can someone help me plz.
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5 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6
yan [13]
Each curve completes one loop over the interval 0\le t\le2\pi. Find the intersections of the curves within this interval.

6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi

The region of interest has an area given by the double integral

\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta

equivalent to the single integral

\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta

which evaluates to 9\pi+72.

8 0
3 years ago
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