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andre [41]
2 years ago
14

Please answer this question!

Mathematics
2 answers:
Archy [21]2 years ago
7 0

Answer:

the answer is 45 . so circle 4

ZanzabumX [31]2 years ago
5 0

Answer:

<em>(4) y = 45</em>

Step-by-step explanation:

\frac{y}{5}  - 2 = 7\\\frac{y}{5} = 7 + 2\\\frac{y}{5}  = 9\\y = 9 (5)\\y =45

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Complete the square for 5x^2 – 30x = 5.
Yuri [45]
First, subtract 5 from both sides, leaving you with 5x^{2} - 30x - 5 = 0
If you use the quadratic formula (a=5), (b=-30), (c=-5)

x=-b +/- √b²-4ac / 2a

x= -(30) +/- √(30)² - 4(5) * (-5) / 2(5)

x= 30 +/- √1000 / 10

x = 3 +/- √10
7 0
3 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
the seating capacity of basketball stadium is 5782 the seats are arranged in 24 sections of the same size any seats over from th
notsponge [240]
5782/24=240 since we round down to s whole.

Multiply again by 24.
240*24=5760

There are 5782-5760=22 priority seats.

The correct answer is 22.
3 0
3 years ago
Determine the maximum number of zeros of the polynomial function
IRINA_888 [86]

Answer:

d

Step-by-step explanation:

6 0
2 years ago
On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
9966 [12]

Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

Circle center (h, k) = (0, 0)

Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

Distance \, formula = \sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}}

Distance² = (x - h)² + (y - k)² = r² which gives;

(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

\sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}} = Distance

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point  (2, √6), does not lie on the circle.

4 0
2 years ago
Read 2 more answers
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