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-2/-9=x/14

2/9=x/14 (\(\frac{2}{9}=\frac{x}{14}\))

cross multiply to get: 9x=28

x=28/9

it is 69, to find it

\(\sqrt{4761}= \sqrt{9*529}=3\sqrt{529}\)

now, we know that 20 is close to 529 and only a units digit of 3 can result in a units digit of 9, so

\(3\sqrt{529}=3\sqrt{23^2}=3*23=69\)

sqrt(4761) = 69.

recheck: 69*69=sqrt(4761)

Which of these statements do you think are correct or incorrect?

We are here to help not to just give ansers.

Please discuss

\(21\times {1\over3} = \color{brown}\boxed{7}\)

You made a mistake Pavlov, in adding \(-4 -10+10 +4 = 0\).

Thus, \(10a+10b+10c+10d = 0\)

So, \(\color{brown}\boxed{a + b + c+d=0}\)

We need to find n such that choosing 3 people from that where order doesn't matter and the number of ways is greater than 365. We use the formula for combination:

ncr = (n!)/r!(n-r)! In this case we will --> (n!)/3!(n-3)!

                          choose 3 people 

The number of ways for this must be greater than 365:

(n!)/3!(n-3)! ≥ 364 --> (n(n-1)(n-2)(n-3)!)/6(n-3)! ≥ 365

                                    n(n-1)(n-2) ≥ 2190
Use trial and error. Use values starting from n = 11:

For n = 15, the number of ways of choosing 3 random people became larger than 2190.

n = 15

\(18\) is incorrect. Here is how you solve it:

\({\text {sum}\over n} = 10\)

\(\text {sum} = 10n\)

\({{10n+50}\over {n+2}} = 12\)

\({10n+50} = 12n+24\)

\(50 = 2n+24\)

\(26 = 2n\)

\(n=13\)

Thus, there are \(\color{brown}\boxed{13}\) terms in the original sequence that make for a mean of \(10\), and when the numbers \(20\) and \(30\) are added, the sum becomes \(180\), making the mean \(12\).

\((6a^3+11a^2-4a-9)\over(3a-2) \)

I thought about this problem as factoring. 

At the end, it will be \((3a-2)(xa^2+ya+z) + r\)

In this equation, \(x\), \(y\), and \(z\) are coeficents, and \(r\) is the remainder.

We know that x must equal 2, because \(2a^2(3a-2)=6a^3-4a^2\)

We know that y equals 5, because (note we already have a -4 so it will add to 11): \(5a(3a-2) = 15a^2-10a\)

We know that z must equal 2, because(Note: we already have a -10, so it will add to -4): \(2(3a-2) = 6a-4\).

When we add everything, we have: \(6a^3+11a^2-4a-4\). This means that the remainder is \(\color{brown}\boxed{-5}\), and the quotient is \(\color{brown}\boxed{2a^2-5a+2}\)

2 * 7 = 14

2 zelks = 14 zacks

14 / 2 * 6

= 7 * 6

= 42

So 14 zacks = 42 zinks

I brute forced the problem. \({{6a^3}\over {3a } } = 2a^2\)... and so on, until I found the quotient, which is the trinomial: \(\color{brown}\boxed{2a^2+5a+2}\).

When you multiply this by \(3a-2\), you get: \({6a^3+11a^2-4a-4}\). The remainder to this problem is \(\color{brown}\boxed{-5}\)

(I'll explain it more in-depth in 45 min once I get home.)

The x-value of the vertex can be also be found by doing \(-b\over2a\), in a quadratic equation with the format \(a^2 + bx + c = 0\).

This also yields \(\color{brown}\boxed5\)

That is much nicer, thanks for reminding me :))

The least exterior angle has to be the greatest interior angle. We know that \(\color{brown}\boxed{\angle{P}}\) is the greatest, because it is greater than the sum of the remaining \(2\) angles. 

There are \(4\) times as many daffodils as roses, which means that there are \(45 \times 4 = 180\) daffodils.

The second peice of information tells us there are \(15\) tulips than roses, meaning there is \(45+15=60\) tulips. 

We already know there are \(45\) roses.

We can not solve anymore, because we do not know how many flowers are in a bouquet.

The red line represents the equation \(y=2x+5\).

The bue point represents the point\((6,2)\)

The green line represents the line that is perpendicular to the red line, while crossing the blue point. 

The reflection must be a point on the green line that has an x-intercept at \(-6\) (\(6\) units to the left of the intersection, since the blue point is \(6\) units to the right)

This means that the reflection is at \(\color{brown}\boxed {(-6,8)}\)

[(0, 2, 2), (2, 0, 2), (2, 2, 0), (2, 2, 2)] >Total distinct permutations = 4

If you accept 0, 2, 2, then there are 4. If not, then there are 3.

m >12

Three.

total possible sequences of flips: 2^10 = 1024 (there are only two choices for each flip, heads or tails, for ten flips)

8 heads: 10 choose 8 = 45 (the number of ways to arrange the sequence HHHHHHHHTT)

9 heads: 10 choose 9 = 10

10 heads: 10 choose 10 = 1

total probability: (45 + 10 + 1)/1024 = 56/1024 = 7/128

(3x+5/6 x )^2  = 9x^2 + 5x + 25/36

oops sorry

View question - quadratic (0calc.com) - answered.