All Answers 352688 Answers

Thank you so much!

Chance of snow 2 days in a row with that info would be:

20% * 60% AKA 0.2 * 0.6

...which is 0.12, or a 12% chance of it snowing both days

How many students in this school?

Happy Birthday IAmJeff!

May all your sweet dreams come true...

Because these two numbers are relatively small, you can easily "guess" that the two numbers can be:

12 and 30, or 6 and 60....etc. If you know one of the the numbers, then you can directly solve for the second.

Happy Birthday to you.

Have a Good, great, and awesome birthday Jeff! Ooh M'Ns! 

asinus,

wow, this is something else!!..would never have thought of going that route. I do appreciate your help! Thank you..

Hi Melody,

a Million thank you's...I think your answer is closest to what I believe the answer should be. I do appreciate!

Here is another way of looking at this problem, which ends with the same result as figured above:

There are 31 primes between 7 and 139 =31[This means that prime numbers from 7 and up can be squared or multiplied by each other to get the desired numbers such as: 7^2, 7*11, 7*13...and so on.

There are 20 primes between 11 and 90=20

There are 16 primes between 13 and 76 =16

There are 10 primes between 17 and 58 =10

There are 8 primes between 19 and 52 =8

There are 6 primes between 23 and 43 =6

There are 2 primes between 29 and 34 =2

There is 1 prime between 31 and 32 =1

That is a total =31 + 20 + 16 + 10 + 8 + 6 + 2 + 1 =94

There is 7^3 for an additional number

There are 7^2*11, 7^2*13, 7^2*17, 7^2*19 = 4 additional numbers.

There is 11^2*7 for 1 additional number.

So, the grand total =94 + 1 + 4 + 1 =100 prime-looking numbers under 1,000.

GYanggg,

thank you for trying...I do appreciate..

Now, let us see what we can do:

There are 999 / 2 =499 numbers that are divisible by 2

There are 999/3 =333 that are divisible by 3. But 166 of them are and are included in 499. But 167 are odd which are not included in 499 above.

So, we have: 499 + 167 =666

We have:999/5=199. But only 100 are odd and end in 5. So, we have:

666 + 100 =766. But, 999/15 =66 half of which are already inculded that end in 5 and are divisible by 3. So, we have: 766 - 33 =733 + 1 itself - 3 primes(2, 3, 5) =731 + 168 primes=899.

So, 999 - 899 = 100 prime-looking numbers under 1,000.

A circle has center O(2, 3) and radius 10. Which of the following points is on the circle?

Hint: \(d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

 V(−3, 2) 

X(5, −2) 

Y(2, 0) 

Z(8, 11)

\(\text{Let $x_c = 2$ } \\ \text{Let $y_c = 3$ } \\ \text{Let $r = 10$ }\)

\(\begin{array}{|lrcll|} \hline & r^2 &=& (x_p - x_c)^2 + (y_p - y_c)^2 \\\\ V(x_p=-3,y_p=2) & 100 &=& (-3 - 2)^2 + (2 - 3)^2 \\ & 100 & \ne & 25 + 1\quad \text{no} \\\\ X(x_p=5,y_p=-2) & 100 &=& (5 - 2)^2 + (-2 - 3)^2 \\ & 100 & \ne & 9 + 25 \quad \text{no} \\\\ Y(x_p=2,y_p=0) & 100 &=& (2 - 2)^2 + (0 - 3)^2 \\ & 100 & \ne & 0 + 9 \quad \text{no} \\\\ Z(x_p=8,y_p=11) & 100 &=& (8 - 2)^2 + (11 - 3)^2 \\ & 100 & = & 36 + 64 \quad \text{yes} \\ \hline \end{array}\)

One more try:

a\(x^{\frac{1}{3}}=4{\frac{1}{x^3}}\)

definition:

\({\color{black}is}\ 4\frac{1}{3}=4 +\frac{1}{3}\ {\color{black}so\ is}\ 4\frac{1}{x^3}=4 +\frac{1}{x^3}\)

\(x^{\frac{1}{3}}=4+{\frac{1}{x^3}}\)         |  \(\times x^3\)

\(x^{\frac{10}{3}}=4x^3+1\)

\(f(x)=4x^3-x^{\frac{10}{3}}+1 \neq 0\)

The function has a minimum in P (0;1).
The function has no zero.

  !

16 weeks x $450 per week =$7,200 - total pay for job A

$1,600  +  $1,840  +  $2,116  +  $2,433.4 =$7,989.4 - total pay for job B

$7,989.40 - $7,200 =$789.40 - extra that the student will earn by accepting job B

Two positive integers $m$ and $n$ are chosen such that $m$ is the smallest positive integer with only two positive divisors and $n$ is the largest integer less than $100$ with exactly three positive divisors. What is $m+n$?

Picture...https://static.k12.com/nextgen_media/assets/8107326-NG_GMT_DP024_232_014.jpg

5. What is m∠V ?

Round only your final answer to the nearest tenth.

53.9º

56.1º

61.9º

70.0º

\(\begin{array}{|rcll|} \hline 7.4^2 &=& 8.6^2+7.6^2-2\cdot 8.6\cdot 7.6\cdot \cos(V) \\ 54.76 &=& 73.96 + 57.76 -130.72\cdot \cos(V) \\ 54.76 &=& 131.72 -130.72\cdot \cos(V) \\ 130.72\cdot \cos(V) &=& 131.72 - 54.76 \\ 130.72\cdot \cos(V) &=& 76.96 \\\\ \cos(V) &=& \dfrac{76.96}{130.72} \\\\ \cos(V) &=& 0.58873929009 \\ \angle{V} &=& 53.9324046479^{\circ} \\ \hline \end{array}\)

\(\angle{V} = 53.9^{\circ} \)

Picture...https://static.k12.com/nextgen_media/assets/8107323-NG_GMT_DP024_232_011.jpg

4. What is the value of g?

Round your answer to the nearest tenth.

18.6 ft

20.4 ft

21.4 ft

24.0 ft

\(\begin{array}{|rcll|} \hline g^2 &=& 25^2+32^2-2\cdot 25\cdot 32\cdot \cos(42^{\circ}) \\ g^2 &=& 1649- 1600\cdot 0.74314482548 \\ g^2 &=& 1649- 1189.03172076 \\ g^2 &=& 459.968279236 \\ g &=& 21.4468710827\ \text{ft} \\ \hline \end{array} \)

\(g = 21.4\ \text{ft}\)

Picture...https://static.k12.com/nextgen_media/assets/8107320-NG_GMT_DP024_232_008.jpg

3. What is the measure of ∠Y ?

Round your answer to the nearest degree.

24°

26°

59°

61°

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(Y) } {35.2} &=& \dfrac { \sin(95^{\circ})} {86} \\ \sin(Y) &=& \dfrac { 35.2\cdot \sin(95^{\circ})} {86} \\\\ \sin(Y) &=& 0.40774480666 \\\\ \angle{Y} &=& 24.0632455972^{\circ} \\ \hline \end{array} \)

\( \angle{Y} = 24^{\circ} \)

The least common multiple of two positive integers is 7!, 

and their greatest common divisor is 9. If one of the integers is 315,

then what is the other?
 

Formula:

\(\begin{array}{|rcll|} \hline GCD(M, N) \times LCM(M, N) &=& M \times N \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline GCD(M, N) \times LCM(M, N) &=& M \times N \\ 9\times 7! &=& 315\cdot N \\\\ N &=& \dfrac{9\times 7!}{315} \\\\ &=& \dfrac{ 7!}{35} \\\\ &=& \dfrac{ 2\times 3 \times 4 \times 5 \times 6 \times 7 }{5\times 7} \\\\ &=& 2\times 3 \times 4 \times 6 \\\\ &=& 144 \\ \hline \end{array} \)

The other is 144

Find the total number of four-digit palindromes.

(Recall that a palindrome is a non-negative sequence of digits which reads the same forwards and backwards, such as 1331. Zero cannot be the first digit.)

There are 90 four-digit palindromes.

\(\begin{array}{|r|r|r|} \hline & n & \text{palindrome } a(n) \\ \hline 1. & 110& 10\ 01 \\ 2. & 111& 11\ 11 \\ 3. & 112& 12\ 21 \\ 4. & 113& 13\ 31 \\ 5. & 114& 14\ 41 \\ 6. & 115& 15\ 51 \\ 7. & 116& 16\ 61 \\ 8. & 117& 17\ 71 \\ 9. & 118& 18\ 81 \\ 10. & 119& 19\ 91 \\ \hline 11. & 120& 20\ 02 \\ 12. & 121& 21\ 12 \\ 13. & 122& 22\ 22 \\ 14. & 123& 23\ 32 \\ 15. & 124& 24\ 42 \\ 16. & 125& 25\ 52 \\ 17. & 126& 26\ 62 \\ 18. & 127& 27\ 72 \\ 19. & 128& 28\ 82 \\ 20. & 129& 29\ 92 \\ \hline 21. & 130& 30\ 03 \\ 22. & 131& 31\ 13 \\ 23. & 132& 32\ 23 \\ 24. & 133& 33\ 33 \\ 25. & 134& 34\ 43 \\ 26. & 135& 35\ 53 \\ 27. & 136& 36\ 63 \\ 28. & 137 & 37\ 73 \\ 29. & 138& 38\ 83 \\ 30. & 139& 39\ 93 \\ \hline 31. & 140& 40\ 04 \\ 32. & 141& 41\ 14 \\ 33. & 142& 42\ 24 \\ 34. & 143& 43\ 34 \\ 35. & 144& 44\ 44 \\ 36. & 145& 45\ 54 \\ 37. & 146& 46\ 64 \\ 38. & 147& 47\ 74 \\ 39. & 148& 48\ 84 \\ 40. & 149& 49\ 94 \\ \hline 41. & 150& 50\ 05 \\ 42. & 151& 51\ 15 \\ 43.& 152& 52\ 25 \\ 44. & 153& 53\ 35 \\ 45. & 154& 54\ 45 \\ 46. & 155& 55\ 55 \\ 47. & 156& 56\ 65 \\ 48. & 157& 57\ 75 \\ 49. & 158& 58\ 85 \\ 50. & 159& 59\ 95 \\ \hline 51. & 160& 60\ 06 \\ 52. & 161& 61\ 16 \\ 53. & 162& 62\ 26 \\ 54. & 163& 63\ 36 \\ 55. & 164& 64\ 46 \\ 56. & 165& 65\ 56 \\ 57. & 166& 66\ 66 \\ 58. & 167& 67\ 76 \\ 59. & 168& 68\ 86 \\ 60. & 169& 69\ 96 \\ \hline 61. & 170& 70\ 07 \\ 62. & 171& 71\ 17 \\ 63. & 172& 72\ 27 \\ 64. & 173& 73\ 37 \\ 65. & 174& 74\ 47 \\ 66. & 175& 75\ 57 \\ 67. & 176& 76\ 67 \\ 68. & 177& 77\ 77 \\ 69. & 178& 78\ 87 \\ 70. & 179& 79\ 97 \\ \hline 71. & 180& 80\ 08 \\ 72. & 181& 81\ 18 \\ 73. & 182& 82\ 28 \\ 74. & 183& 83\ 38 \\ 75. & 184& 84\ 48 \\ 76. & 185& 85\ 58 \\ 77. & 186& 86\ 68 \\ 78. & 187& 87\ 78 \\ 79. & 188& 88\ 88 \\ 80. & 189& 89\ 98 \\ \hline 81. & 190& 90\ 09 \\ 82. & 191& 91\ 19 \\ 83. & 192& 92\ 29 \\ 84. & 193& 93\ 39 \\ 85. & 194& 94\ 49 \\ 86. & 195& 95\ 59 \\ 87. & 196& 96\ 69 \\ 88. & 197& 97\ 79 \\ 89. & 198& 98\ 89 \\ 90. & 199& 99\ 99 \\ \hline \end{array}\)

If k is constant, what is the value of k so that the polynomial

\(k^2x^3-8kx+16\) is divisible by \(x-1\)?

\(k^2x^3-8kx+16 = k^2(x-1)(\ldots )(\ldots) \)

The first root is 1.

\(\begin{array}{|rcll|} \hline k^2 \cdot 1^3 - 8k \cdot 1 + 16 &=& 0 \\ k^2 - 8k + 16 &=& 0 \\ k &=& \dfrac{ 8\pm \sqrt{64-4\cdot 16} }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ k &=& \dfrac{ 8\pm 0 }{2} \\ \mathbf{k} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 16x^3-32x+16 : (x-1) = 16x^2+16x-16 \\ \hline \end{array}\)