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Sorry, but how is having a "x" angle the same as 180-x ?
(If x=60 then it is acute, but 180-60 = 120 is obtuse?)

It is correct since the angle could be \(180^\circ - \cos^{-1}\left(\dfrac16\right)\). Having a \(x^\circ\) angle is basically the same thing as having a \((180 - x)^\circ\) angle.

-7y+14, expanding using the distributive property

~ saph

Very similar solved problem here:

https://web2.0calc.com/questions/product_4

Use sine law directly, \(\dfrac{BC}{\sin 43^\circ} = \dfrac{12\,\text{ m}}{\sin 99^\circ}\)

Please continue from here.

Using sine law directly gives \(\dfrac{AC}{\sin 58^\circ} = \dfrac{86}{\sin 84^\circ}\).

Can you take it from here? It is already one step from the answer.

\(\dfrac{7}{\sin B} = \dfrac 6{\sin 30^\circ}\\ 6\sin B = 7\sin 30^\circ\\ 6 \sin B = \dfrac72\)

Now this is a linear equation in sin B, so you can solve for sin B by dividing by 6 on both sides.

No matter B is acute or not, sin B is the same.

The number of such matches equals the number of ways that you can choose 2 left-handed person from 5 of them to play a match, which is \(\displaystyle\binom{5}{2} = 10\).

You can use cosine law to find c first.

\(c^2 = 8^2 + 11^2 - 2\cdot 8 \cdot 11 \cos 37^\circ\\ c \approx 6.666344593\)

Then, use cosine law again to find cos A.

\(8^2 = c^2 + 11^2 - 2\cdot c \cdot 11 \cos A\)

Substitute the value of c calculated above. Can you take it from here?

Answered here: https://web2.0calc.com/questions/number-theory_23228

Let x be the minimum length of CQ. Note that by Power of Points formula, AQ * BQ = CQ * DQ.

\(AQ \cdot BQ = CQ \cdot DQ\\ 16 \cdot 12 = x \cdot (36 - x)\\ x^2 - 36x + 192 = 0\\ x = 18 - 2 \sqrt {33}\text{ or }x = 18 + 2\sqrt {33}\)

The smaller one is 18 - 2 * sqrt(33), and that's the answer.

Answered here: https://web2.0calc.com/questions/please-help_87485#r1

Notice that for any 2 consecutive terms in the product, the numerator of the first fraction always cross out with the denominator of the second fraction. That leaves us with the numerator of the last fraction, divided by the denominator of the first fraction, which is (n + 20)/n. If you are attempting the solve the equation, you should try to multiply both sides by n so that the equation becomes linear.

You know that 

\(a_1 = 48 - 3\times1 = 45\\ d = a_2 - a_1 = (48 - 3(2)) - (48 - 3(1)) = -3\\ \)

where d is the common difference of the sequence.

The sum of A.S. formula states \(a_1 + a_2 + \cdots + a_n = \dfrac n2 \left(2a_1 + (n - 1)(a_2 - a_1)\right) = \dfrac n2 (2a_1 + (n - 1)d)\) if a_n is an arithmetic sequence. Therefore, the equation \(a_1 + a_2 + a_3 + \cdots + a_k = 330\) really says \(\dfrac k2 (2(45) + (k - 1)(-3)) = 330\). This becomes a quadratic equation. Can you solve for the value of k now?

Hint: Expand and use quadratic formula.

You can let \(f(x) = x^4 + ax^3 + bx^2 + cx + d\) be the polynomial. You can then use the given information to find the missing coefficients.

Substituting x = -1, 2, -3, 4 respectively gives \(\begin{cases}1-a+b-c+d = -1\\16 + 8a + 4b + 2c + d = -4\\81 - 27a + 9b - 3c + d = 9\\256 + 64a + 16b + 4c + d = -16\end{cases}\).

Solving gives: 

So f(1) = \(1 + \left(-\dfrac{79}{35}\right) + \left(-\dfrac{89}7\right) + \dfrac{432}{35} + \dfrac{768}{35}\) = 711/35.

This is the diagram.

Note that \(\angle BPC = 180^\circ - 116^\circ = 64^\circ\).

Since BC = CP, \(\angle PBC = 64^\circ\).

Considering the interior angle sum of \(\triangle BPC\) gives \(\angle PCB = 52^\circ\).

Since AB = AC, \(\angle ACB = \angle ABC = \angle PBC = 64^\circ\).

Therefore, \(\angle ACP = \angle ACB - \angle PCB = 64^\circ - 52^\circ = 12^\circ\).

Consider the prime factorization of n. Suppose that \(n = 2^x 3^y 5^z\). (It does not have any other prime factors because n is the smallest one)

Then \(2n = 2^{x + 1} 3^y 5^z\), \(3n = 2^{x} 3^{y + 1} 5^z\), and \(5n = 2^x 3^y 5^{z + 1}\).

2n being a perfect square means all exponents are divisible by 2, i.e., x is odd, y and z are even.

3n being a perfect cube means all exponents are divisible by 3, i.e., x, y + 1, and z are divisible by 3.

5n being a perfect fifth power means all exponents are divisible by 5, i.e., x, y, and z + 1 are divisible by 5.

The smallest odd value of x that is divisible by 3 and 5 is 15.

The smallest even value of y such that y + 1 is divisible by 3 and y is divisible by 5 is 20.

The smallest even value of z such that z is divisible by 3 and z + 1 is divisible by 5 is 24.

Therefore, \(n = 2^{15}3^{20}5^{24}\) is the required answer.

Since the coefficients are rational and the roots are irrational, the roots are \(x = 1\pm \sqrt 2\) and \(x = 2\pm \sqrt 7\).

The quartic polynomial required is 

\(\quad(x - (1 + \sqrt 2))(x - (1 - \sqrt 2))(x - (2 + \sqrt 7))(x - (2 - \sqrt 7))\\ = (x^2 - 2x - 1)(x^2 - 4x - 3)\)

Can you expand that on your own?

1024

Graph y=(3x+4)+(x^2 - 5 + 6) + (|3-x|-8) + (((x+4)^3) - 8) + ((sqrt(5-x))-3) + (log_2(x) - 4) - (1/x) + (x^2-8+15)/(x-3)(x+4) - 3^x + 9

Did you have fun making that up ?

Find AC.

Hello Guest!

Law of cosines:

\((7\sqrt{3} )^2=9^2+x^2-2\cdot9\cdot x\cdot cos\ 150°\\ x^2-cos\ 150°\cdot18x+81-147=0\\ x=cos\ 150°\cdot 9\pm \sqrt{81\cdot cos^2\ 150°+147}\\ x=-7.79423\pm \sqrt{207.75}\\ x\in \{{\color{blue}6.6193},-22.207\}\)

\(\color{blue}\overline{AC}=6.6193\)

  !

M has x

W has 7/13x

P give w y stamps and M y stamps.

So after that, M has x + y & w has 7/13x + y

 and 7/13x + y = (x + y) 9/10, 500 < y < 550

               7/13x + y = 9/10x + 9/10y

                     1/10y = 47/130x

                            y = 47/13x

The important is x, y lol are positive integers.

coz 500 < y < 550 => 500 < 47/13x < 550

                                   138.30 < x < 152.13

By analyzing, x should be removed by 13

13 * 10 = 130; 13 * 11 = 143; 13 * 12 = 156. Only 138 < 143 < 152

So x = 143 => y = 517

All in all P give M 517 stamps.

By LOS (Law of Sines), AC = 8*sqrt(3)/3.

We have the equation: \(x^2 + y^2 - 9 = 0\)

Adding 9 to both sides gives us: \(x^2 + y^2 = 9 \), where 9 is the radius squared. 

Can you take it from here?

5) Using the Pythagorean Theorem on \(\triangle ADC\), we find that \(\overline{AD} = 8\)

Now, applying the Pythagorean Theorem on \(\triangle ADB\), we find that \(\overline{AB} = 17 \).

We also know that \(\sin = {\text{opposite} \over \text{hypotenuse}}\), which in this case, is \({\overline{AD}} \over {\overline{AB}}\).

Can you take it from here?