All Answers 356462 Answers

the domain is the set of all the x values

the range is the set of all the y values.

\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C &= \left( \frac{1}{8^{1/5}} \right)^2\\ D &= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E &= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}\vphantom{dfrac{2}{2}} \end{align*}

\(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}=2^{0.5-0.\dot3}=2^{1/6}=2^{2/12}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}=\sqrt[12]{2^7}\vphantom{dfrac{2}{2}}=2^{7/12} \end{align*} etc\)

Change them all to powers of 2.

And stop accusing other people of cheating (on other threads)

It is irritating.

\(2f(4)=2(16-8+m)=2(8+m)\\ g(4)=16-8+4m=8+4m\\~\\ 2(8+m)=8+4m\\ etc \)

Another answer here:

What about 100?  I think you have missed many others as well.

if the middle digit is 0 the last digit can be any                      10

if the middle digit is 1 the last digit can be 2 or bigger            8

if the middle digit is 2 the last digit can be 4 or bigger            6

if the middle digit is 3 the last digit can be 6 or bigger            4

if the middle digit is 5 the last digit can be 8 or bigger            2

Total so far = 30

the first digit can be 1 to 9    

30*9=270

What is the slope of the line containing the midpoint of the segment with endpoints at (2 , 4) and (0 , -2) and the midpoint of the segment with endpoints at  (5 , 1) and (1 , 5) Express your answer in simplest form.

This is amazing:  r = ( UY / 2 )²  

So far I have only this:  r = 65

And now I have:   WX = 2*[sqrt( 65² - 56² )] = 66  

If you wait too long the thread might be locked.

In which case I think a padlock is displayed, ...at the top I think.

When you make comments like this it is a good idea for you to include a link to the post where it has happened.

Then we might be able to see why it happened.

Maybe this pic will help you to understand.

Hi,

thanks for showing your effort.

Here is my answer.

\(3[log(x + 1) + log(x + 4)]\\ =3log[(x + 1)(x + 4)]\\ =log[(x + 1)(x + 4)]^3\\ \)

Yes, our answers are the same.  Good work :)

Two numbers, if the first one increases by 1, and the second one decreases by 1, then their product increases by 2020.
If the first number decreases by 1, and the second one increases by 1, what value does the product decrease?

\(\text{ Let first number $=n_1$} \\ \text{ Let second number $=n_2$}\)

\(\begin{array}{|lrcll|} \hline (1) & (n_1+1)(n_2-1) &=& n_1n_2 + 2020 \\ & n_1n_2-n_1+n_2-1 &=& n_1n_2 + 2020 \\ & -n_1+n_2-1 &=& 2020 \\ & \mathbf{ -n_1+n_2 } &=& \mathbf{ 2021 } \\ \hline (2) & (n_1-1)(n_2+1) &=& n_1n_2 + x \\ & n_1n_2+n_1-n_2-1 &=& n_1n_2 + x \\ & n_1-n_2-1 &=& x \\ & \mathbf{ n_1-n_2-1 } &=& \mathbf{ x } \\ \hline (1)+(2): & -n_1+n_2 + n_1-n_2-1 &=&2021 +x \\ & -1 &=&2021 +x \\ & x &=& -2021 -1 \\ & \mathbf{x} &=& \mathbf{-2022} \\ \hline \end{array}\)

The value of the product decreases by 2022

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius 1.
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

Thanks for asking Qwerty.

Click on the chain link in the ribbon. It is on the bottom left.

And copy your link in.  You can copy it once for the display and once for the link.

Experiment and you will work it out very quickly.

Let \(f(x)\)  be a polynomial such that \(f(0) = 4\), \(f(1) = 5\), and \(f(2)=10\).
Find the remainder (=r) when \(f(x)\) is divided by \(x(x - 1)(x - 2)\).

Chords \(\overline{UV}\), \(\overline{WX}\), and \(\overline{YZ}\) of a circle are parallel.
The distance between chords \(\overline{UV}\) and \(\overline{WX}\) is 4,
and the distance between chords \(\overline{WX}\) and \(\overline{YZ}\) is also 4.
If \(\overline{UV}\) = 78 and \(\overline{YZ}\) = 50, then find \(\overline{WX}\).

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{r} &=& \mathbf{a+b+8} \\ \hline & \mathbf{39^2} &=& \mathbf{(b+8)(a+r)} \\ & 39^2 &=& (b+8)(a+a+b+8) \\ (2) & \mathbf{39^2} &=& \mathbf{(b+8)(2a+b+8)} \\ \hline & \mathbf{25^2} &=& \mathbf{b(8+a+r)} \\ & 25^2 &=& b(8+a+a+b+8) \\ (3) & \mathbf{25^2} &=& \mathbf{b(2a+b+16)} \\ \hline & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (b+4)(4+a+r) \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (b+4)(4+a+a+b+8) \\ (4) & \mathbf{\left(\dfrac{\overline{WX}}{2}\right)^2} &=& \mathbf{(b+4)(2a+b+12)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3): & \mathbf{25^2} &=& \mathbf{b(2a+b+16)} \\ & 625 &=& b(2a+b+16) \\ & 625 &=& 2ab+b(b+16) \\ & 2ab &=& 625- b(b+16) \\ & \mathbf{a} &=& \mathbf{\dfrac{625- b(b+16)}{2b}} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2): & \mathbf{39^2} &=& \mathbf{(b+8)(2a+b+8)} \\ & 1521 &=& (b+8)(2a+b+8) \quad | \quad \mathbf{a=\dfrac{625- b(b+16)}{2b}} \\ & 1521 &=& (b+8)\left(2\dfrac{\Big(625- b(b+16)\Big)}{2b}+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{ 625- b(b+16) }{b}+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{625}{b}- (b+16)+b+8 \right) \\ & 1521 &=& (b+8)\left(\dfrac{625}{b}-8 \right) \\ & 1521 &=& (b+8)\dfrac{(625-8b)}{b} \\ & 1521b &=& (b+8)( 625-8b) \\ & 1521b &=& 625b-8b^2+8*625-64b \\ & 8b^2 + 960b - 5000 &=& 0 \quad | \quad : 8 \\ & \mathbf{b^2 + 120b - 625} &=& \mathbf{0} \\ \hline &b &=& \dfrac{-120\pm \sqrt{120^2-4*(-625) } }{2} \\ &b &=& \dfrac{-120\pm \sqrt{16900} }{2} \\ &b &=& \dfrac{-120\pm 130 }{2} \\\\ &\mathbf{b} &=& \mathbf{\dfrac{-120+ 130 }{2}} \\ &\mathbf{b} &=& \mathbf{5} \\\\ &\mathbf{b} &=& \mathbf{\dfrac{-120- 130 }{2}} \quad | \quad \text{no solution } b < 0!\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{\dfrac{625- b(b+16)}{2b}} \quad | \quad \mathbf{b=5} \\\\ a &=& \dfrac{625- 5(5+16)}{2*5} \\\\ a &=& \dfrac{625- 5*21}{10} \\\\ a &=& \dfrac{625- 105}{10} \\\\ a &=& \dfrac{520}{10} \\\\ \mathbf{a} &=& \mathbf{52} \\ \hline \end{array}\)

\(\mathbf{\overline{WX}=\ ?}\)

\(\begin{array}{|rcll|} \hline (4): & \mathbf{\left(\dfrac{\overline{WX}}{2}\right)^2} &=& \mathbf{(b+4)(2a+b+12)} \quad | \quad a=52,\ b=5 \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& (5+4)(2*52+5+12) \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& 9*121 \\ & \left(\dfrac{\overline{WX}}{2}\right)^2 &=& 1089 \\\\ & \dfrac{\overline{WX}}{2} &=& \sqrt{1089} \\\\ & \dfrac{\overline{WX}}{2} &=& 33 \\\\ &\mathbf{\overline{WX}} &=& \mathbf{66} \\ \hline \end{array}\)

cos(5º) = 0.996194698

sin(25º) = 0.422618261

sin(θ) = 0.996194698 - 0.422618261 = 0.573576436

Angle  θ = 35º

cos(5º) = 0.996194698

sin(25º) = 0.422618261

sin(θ) = 0.996194698 - 0.422618261 = 0.573576436

Angle  θ = 35º

(102, 103, 104, 105, 106, 107, 108, 109, 112, 113, 114, 115, 116, 117, 118, 119, 124, 125, 126, 127, 128, 129, 136, 137, 138, 139, 148, 149, 202, 203, 204, 205, 206, 207, 208, 209, 212, 213, 214, 215, 216, 217, 218, 219, 224, 225, 226, 227, 228, 229, 236, 237, 238, 239, 248, 249, 302, 303, 304, 305, 306, 307, 308, 309, 312, 313, 314, 315, 316, 317, 318, 319, 324, 325, 326, 327, 328, 329, 336, 337, 338, 339, 348, 349, 402, 403, 404, 405, 406, 407, 408, 409, 412, 413, 414, 415, 416, 417, 418, 419, 424, 425, 426, 427, 428, 429, 436, 437, 438, 439, 448, 449, 502, 503, 504, 505, 506, 507, 508, 509, 512, 513, 514, 515, 516, 517, 518, 519, 524, 525, 526, 527, 528, 529, 536, 537, 538, 539, 548, 549, 602, 603, 604, 605, 606, 607, 608, 609, 612, 613, 614, 615, 616, 617, 618, 619, 624, 625, 626, 627, 628, 629, 636, 637, 638, 639, 648, 649, 702, 703, 704, 705, 706, 707, 708, 709, 712, 713, 714, 715, 716, 717, 718, 719, 724, 725, 726, 727, 728, 729, 736, 737, 738, 739, 748, 749, 802, 803, 804, 805, 806, 807, 808, 809, 812, 813, 814, 815, 816, 817, 818, 819, 824, 825, 826, 827, 828, 829, 836, 837, 838, 839, 848, 849, 902, 903, 904, 905, 906, 907, 908, 909, 912, 913, 914, 915, 916, 917, 918, 919, 924, 925, 926, 927, 928, 929, 936, 937, 938, 939, 948, 949) = Total 252

Let the original of money =M
1/2M + 10 =She spent at the first store.
M - [1/2M+10]=[M - 20]/2 -what she had left after the first store, 
 (M - 20)/2 =1/2(M - 20)/2 + 10, solve for M
M = $60 - The original amount of money she started with.
Check: $60/2 + 10 =$40 what she spent at first store.
$60 - $40 =$20 she had left after the first store.
$20 / 2 + $10 =$20 what she spent at the 2nd store.
$20 - $20 =0 - what she had left after the 2nd store.

Thanks the hint helped a lot!

we have -1^2-2*-1=1--2=3 so f(-1)=3

Quiz scores for a group are 50, 75, 70, 80, 100, x, and y. Find values for x and y so that all the following conditions are met: 

1/     the mode is 70

2/     the median score is 75

3/     the mean score is 77

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Scores are:   50, 75, 70, 80,100, x, and y

Mode //   If the mode is 70, either x or y must be 70.      x = 70

Mean //    (50+70+70+75+ 80 +100+y) / 7 = 77       y = 94

Median //   50,70,70,75,80,94,100