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so we simplify the left side of the equation to get −x^2+10x=40 or 10x-x^2 we then subtract 40 from both sides to get −x^2+10x−40=0 and then we use the quadratic formula to get x=−10+sqrt(−60)−2 or x=−10-sqrt(−60)−2

We have a pattern 

and then we divide by 4 to get 29/4=7 R1 where R means the remainder so it is the first one 8

Here's an alternative approach:

Find the coefficient of $y^4$ in the expansion of $(2y - 5)^7.$

\(\begin{pmatrix} 7\\ 4 \end{pmatrix} (2y)^4(-5)^3\)

you can finish it

Find the sum of length, breadth & height of a cuboid whose whole surface area is 214 cm^2, with a base area of 42 cm^2, and with volume 210 cm^3.

You can't have cosine of an angle greater than 1.  Did you mean cos Z = 5/7?

The side length of the cube will equal the diameter of the ball.  The diameter of the ball is given by circumference/pi.

i.e.  d = 264/pi cm

Can you take it from here?

https://web2.0calc.com/questions/hello-plz-help-the-answers-i-ve-received-so-far-are#r3

5+6+7 = sqrt(18)² 

If  \(x + \dfrac{1}{x} = 1\),  then find the value of  \(x^{1000} + \dfrac{1}{x^{1000}}\)

\(\text{Let $x=e^{i\varphi} $} \\ \text{Let $\dfrac{1}{x}=x^{-1}=e^{-i\varphi}=e^{i(-\varphi)} $} \\ \text{Let $x^{1000}=e^{i(1000\varphi)} $} \\ \text{Let $\dfrac{1}{x^{1000}}=x^{-1000}=e^{-i\varphi 1000}=e^{i(-1000\varphi)} $}\)

\(\begin{array}{|rcll|} \hline x + \dfrac{1}{x} &=& 1 \\ x + x^{-1} &=& 1 \\ e^{i\varphi} + e^{i(-\varphi)} &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(-\varphi)+i\sin(-\varphi) &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(\varphi)-i\sin(\varphi) &=& 1 \\ 2\cos(\varphi) &=& 1 \\ \cos(\varphi) &=& \dfrac{1}{2} \\ \varphi &=& \arccos\left( \dfrac{1}{2} \right) \\ \mathbf{ \varphi } &=& \mathbf{60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{x^{1000} + \dfrac{1}{x^{1000}}} \\ &=& e^{i(1000\varphi)} + e^{i(-1000\varphi)} \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(-1000\varphi)+i\sin(-1000\varphi) \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(1000\varphi)-i\sin(1000\varphi) \\ &=& 2\cos(1000\varphi) \quad | \quad \mathbf{ \varphi=60^\circ} \\ &=& 2\cos(1000*60^\circ) \\ &=& 2\cos(60000^\circ) \\ &=& 2\cos(240^\circ) \\ &=& 2(-0.5) \\ &=& \mathbf{-1} \\ \hline \end{array}\)

If \(\dfrac{\sin^4 x}2 + \dfrac{\cos^4 x}3 = \dfrac15 \) ,  find  \(tan^2x.\)

Hello Guest!

\(\frac{1}{2}(sin^2x)^2+\frac{1}{3}(cos^2x)^2=\frac{1}{5}\\ \frac{1}{2}(sin^2x)^2+\frac{1}{3}(1-sin^2x)^2=\frac{1}{5}\)

\(sin^2x =u\)

\(\frac{1}{2}u^2+\frac{1}{3}(1-u)^2=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}(1-2u+u^2)=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}-\frac{2}{3}u+\frac{1}{3}u^2=\frac{1}{5}\)

\(\frac{1}{2\cdot 3\cdot 5}\cdot (15u^2+10-20u+10u^2)=\frac{6}{2\cdot 3\cdot 5}\)

\(25u^2-20u+4=0\)

\(u = {20 \pm \sqrt{20^2-4\cdot 25\cdot 4} \over 2\cdot 25}\\ u=\frac{1}{50}\cdot (20\pm 0)\\ \color{blue}u=0.4\)

\(sin^2x=\frac{tan^2x}{1+tan^2x}\)

\(tan^2x=v\)

\(\frac{v}{1+v}=u\\ v=u+uv\\ v-uv=u\\ \color{blue}v=\frac{u}{1-u} \)

\(tan^2x=\frac{0.4}{1-0.4}\)

\(tan^2x=\frac{2}{3}\)

Thanks heureka!

  !

If we divide a 2-digit positive integer by the sum of its digits,
we get the quotient and remainder of 4 and 3, respectively.
If we divide the same 2-digit positive by the product of its digits,
we get quotient and remainder of 3 and 5, respectively.
What is the 2-digit integer?

The 2-digit integer \(ab\) is \(10a + b\)

\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the sum of its digits} \\ \end{array} \\ \hline (1): & \mathbf{\dfrac{10a + b}{a+b}} &=& \mathbf{4 + \dfrac{3}{a+b}} \quad | \quad \times (a+b) \\\\ & 10a + b &=& 4(a+b) + 3 \\ & 10a + b &=& 4a+4b + 3 \\ & 10a-4a + b-4b &=& 3 \\ & 6a-3b &=& 3 \quad | \quad : 3 \\ & 2a-b &=& 1 \\ & \mathbf{b} &=& \mathbf{2a-1} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the product of its digits} \\ \end{array} \\ \hline (2): & \mathbf{\dfrac{10a + b}{ab}} &=& \mathbf{3 + \dfrac{5}{ab}} \quad | \quad \times (ab) \\\\ & 10a + b &=& 3ab+5 \quad | \quad \mathbf{b=2a-1} \\ & 10a + 2a-1 &=& 3a(2a-1) + 5 \\ & 12a-1 &=& 6a^2-3a + 5 \\ & 6a^2 -15a +6 &=& 0 \quad | \quad : 3 \\ & \mathbf{2a^2 -5a +2} &=&\mathbf{ 0 } \\\\ & a &=& \dfrac{5\pm \sqrt{5^2-4*2*2} }{2*2} \\ & a &=& \dfrac{5\pm \sqrt{9}}{4} \\ & a &=& \dfrac{5\pm 3}{4} \\\\ & a &=& \dfrac{5+ 3}{4} \\\\ & a &=& \dfrac{8}{4} \\\\ & \mathbf{a} &=& \mathbf{2} \\\\ \text{or} & a &=& \dfrac{5- 3}{4} \\\\ & a &=& \dfrac{2}{4} \qquad \text{no integer }! \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{2a-1} \quad | \quad \mathbf{a=2} \\\\ b &=& 2*2-1 \\ \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}\)

The 2-digit integer ab is \(\mathbf{23}\)

AB = 10         BC = AD = 14            angle A = 45º

h = sin(45º) * 10 = 7.071

Area = h * BC = 98.995 units2      or      (10 * 14) / √2 = 98.99494937  

The units digit is 1.

From the formula, the area of the parallelogram is 1/2*10*14 = 70.

Thanks!! 

We first put one ball in each box. Now we have 2 balls to distribute in 4 boxes. Since they are both identical, the cases are

3 1 1 1

2 2 1 1

So there are 2 possible distributions.

Volume of a cylinder   =  pi r ^2   h         (h is the length   and r = 1 cm in this example)

160       =    pi   1^2   h

160/pi = h = length = ..........      cm

I solved these!

1) The area of triangle ABC is 72*sqrt(2).

2) By similar triangles, PQ = 12*sqrt(3).

Let d be the number of dollars that Jeri finds.  Then solving the inequalities, you get d >= 200 and d < 575, so the answer is [200,575).

From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, B - A = -9/19 - 3/19 = -12/19.

Hello, this has been posted before. 

Solve for x:
x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) = 179

Rewrite the left hand side by combining fractions. x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) = 1/4 (4 x - sqrt(2) sqrt(7 + 4 x - sqrt(8 x + 13))):
1/4 (4 x - sqrt(2) sqrt(7 + 4 x - sqrt(8 x + 13))) = 179

Multiply both sides by 4:
4 x - sqrt(2) sqrt(7 + 4 x - sqrt(8 x + 13)) = 716

Subtract 4 x from both sides:
-sqrt(2) sqrt(7 + 4 x - sqrt(8 x + 13)) = 716 - 4 x

Raise both sides to the power of two:
2 (7 + 4 x - sqrt(8 x + 13)) = (716 - 4 x)^2

2 (7 + 4 x - sqrt(8 x + 13)) = 14 + 8 x - 2 sqrt(8 x + 13):
14 + 8 x - 2 sqrt(8 x + 13) = (716 - 4 x)^2

Subtract 8 x + 14 from both sides:
-2 sqrt(8 x + 13) = -14 + (716 - 4 x)^2 - 8 x

Raise both sides to the power of two:
4 (8 x + 13) = (-14 + (716 - 4 x)^2 - 8 x)^2

Expand out terms of the left hand side:
32 x + 52 = (-14 + (716 - 4 x)^2 - 8 x)^2

Expand out terms of the right hand side:
32 x + 52 = 256 x^4 - 183552 x^3 + 49306240 x^2 - 5881029024 x + 262801820164

Subtract 256 x^4 - 183552 x^3 + 49306240 x^2 - 5881029024 x + 262801820164 from both sides:
-256 x^4 + 183552 x^3 - 49306240 x^2 + 5881029056 x - 262801820112 = 0

The left hand side factors into a product with four terms:
-16 (2 x - 377) (2 x - 339) (4 x^2 - 1436 x + 128519) = 0

Divide both sides by -16:
(2 x - 377) (2 x - 339) (4 x^2 - 1436 x + 128519) = 0

Split into three equations:
2 x - 377 = 0 or 2 x - 339 = 0 or 4 x^2 - 1436 x + 128519 = 0

Add 377 to both sides:
2 x = 377 or 2 x - 339 = 0 or 4 x^2 - 1436 x + 128519 = 0

Divide both sides by 2:
x = 377/2 or 2 x - 339 = 0 or 4 x^2 - 1436 x + 128519 = 0

Add 339 to both sides:
x = 377/2 or 2 x = 339 or 4 x^2 - 1436 x + 128519 = 0

Divide both sides by 2:
x = 377/2 or x = 339/2 or 4 x^2 - 1436 x + 128519 = 0

Divide both sides by 4:
x = 377/2 or x = 339/2 or x^2 - 359 x + 128519/4 = 0

Subtract 128519/4 from both sides:
x = 377/2 or x = 339/2 or x^2 - 359 x = -128519/4

Add 128881/4 to both sides:
x = 377/2 or x = 339/2 or x^2 - 359 x + 128881/4 = 181/2

Write the left hand side as a square:
x = 377/2 or x = 339/2 or (x - 359/2)^2 = 181/2

Take the square root of both sides:
x = 377/2 or x = 339/2 or x - 359/2 = sqrt(181/2) or x - 359/2 = -sqrt(181/2)

Add 359/2 to both sides:
x = 377/2 or x = 339/2 or x = 359/2 + sqrt(181/2) or x - 359/2 = -sqrt(181/2)

Add 359/2 to both sides:
x = 377/2 or x = 339/2 or x = 359/2 + sqrt(181/2) or x = 359/2 - sqrt(181/2)

x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) ⇒ 339/2 - sqrt(7/8 - sqrt(13/64 + 339/(8 2)) + 1/2×339/2) = 321/2:
So this solution is incorrect

x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) ⇒ 377/2 - sqrt(7/8 - sqrt(13/64 + 377/(8 2)) + 1/2×377/2) = 179:
So this solution is correct

x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) ≈ 160.974:
So this solution is incorrect

x - sqrt(7/8 - sqrt(x/8 + 13/64) + x/2) ≈ 179.5:
So this solution is incorrect

The solution is:

 x = 377/2