All Answers 356356 Answers

Number of Solid beads = 26.66

Number of Opaque beads = 16

Number of transparent beads = 5.33

Total number of beads = 48

Ratio of components = 5 : 3 : 1

Ratio of solid beads \(= {5 \over 5 + 3 + 1} = \)\( {5 \over 9}\frac{}{}\)

Ratio of opaque beads = \({3\over 9}\frac{}{}\)

Ratio of transparent = \( {1\over 9}\frac{}{}\)

: Number of Solid beads used = 48 *  \({5 \over 9}\frac{}{}\)= \( {80\over 3}\frac{}{}\)= 26.66

Number of opaque beads used = 48 * \( {3{} \over 9}\frac{}{}\)= 16

Number of transparent beads used = 48 * \( {1 \over 9}\frac{}{}\)=\({16 \over 3}\frac{}{}\)= 5.33

My answer is 79.238 m²

(square area - area of the 30 degree sector)

(96 -16.762 = 79.238)

1) Area of a sector of circle =  (area of a circle )* (Degree angle of the sector / 360)

..[30/360 = 1/12 is like the sector is one part of the whole  12 parts of the whole area of (pi r ² )]...

(pi * 8²  ) * (1/12)  =  16.7619m²

2) The dry area of the garden :  Garden area  - 16.7619  =   79.238 m²

Amazon.com has a multitude of them.

Do you mean graphing with desmos?

The smallest integer on the list is 12345

Second smallest integer is 12354

The first 24 smallest integers is 1---- (since 4!=24)

The next 24 smallest integers are (2----) (again, since 4!=24)

The next 24 smallest integers are (3-----) (since 4!=24)

However, I have already counted 72 integers. The question asks for the 60th integer. So, instead of the ten-thousand's place, I will go for the thousand's place. 

I already know that the integer is between 25431 (the largest integer of the list that starts with 2, this is the 48th smallest integer)

and 35421 (the largest integer that starts with 3 this is the 72nd smallest integer)

I can further narrow it down , since I know that 31--- is 6 integers (since 3!=6), and so that's 48+6=54 integers. 

I know that 32--- is another 6 integers (again, since 3!=6), so once again, that's another 6 integers. 

32541 is the largest integer in the list of the form 32---. So, 32541 is the 60th integer of the list. 

JP

I'm not sure if this is the most efficient way, but you have to start by finding the perpendicular line to y = 3x + 2 that has (5, 3)

y = -1/3x + 14/3

y = -1/3x + 14/3 and y = 3x + 2 intersect at (.8, 4.4)

If (.8, 4.4) is the center point, and (5, 3) is the other point, then the final point is (-3.4, 5.8)

I'm sorry if that was confusing. 

=^._.^=

y=ax^2+bx+c

first substitute (-1, 8) (1 for x and 8 for y)

8=a-b+c

then substitute (2, -4) (2 for x and -4 for y)

-4=4a+2b+c

then substitute (-6, -12) (-6 for x and -12 for y)

-12=36a-6b+c

now we have these three equations:

8=a-b+c

-4=4a+2b+c

-12=36x-6x+c

so a=-1

b=-3

and c=6

and the quadratic equation's equation is y=-x^2-3x+6

JP

Let the dimensions of the package be \(a \) by \(b\) by \(c \).  Thus, the volume is \(abc\). 

The total material required to create the package would be the surface area, namely, \(2(ab+bc+ac)\).  With this, we seek to find an inequality that links \(ab+bc+ac\) to \(abc\), and the AM-GM Inequality does just that.  

By the AM-GM Inequality for 3 variables, \(\frac{ab+bc+ac}{3}≥\sqrt[3]{abc^2}\), so \(\frac{ab+bc+ac}{3}≥\sqrt[3]{1440000}\).  

Simplifying, we have \(ab+bc+ac≥60\sqrt[3]{180}\).  The minimum namely, the right side of the above equation is achieved when the equality condition of the AM-GM is met, or when ab=bc=ac.  This simplifies to a=b=c, or a=b=c= \(2\sqrt[3]{150}\), from abc = 1200

My answer for the dimensions would be \(2\sqrt[3]{150}\) by \(2\sqrt[3]{150}\) by \(2\sqrt[3]{150}\).

*** edit ****

We need dimensions that are in whole number integers to fit the cubes in the box

10 x 10 x 12    would work      surface area   = 680 cm²

at 45 degrees and 225 degrees   sin and cos are equal and of the the same +- sign      Q 1   Q 3

2:7   means the line can be divided into   2+7 = 9 segments

from   3 to -4   is    -7         2 out of 9 segments would be    -7 * 2/9 added to  3 = 13/9 = 1 4/9 = x

          6 to 3    is -3            2 out of 9 segments would be     -3 * 2/9 added to 6  = 5 1/3 = y

Prime factorization 336 is 2^4*3*7.

7 must be one of the digits. The last four digits must be: (7, 6, 4, 2), (7, 6, 8, 1), (7, 3, 8, 2), or (7, 3, 4, 4). Each of the first three produces 4!=24. 24*3=72. The last one produces 3!=12 posssibilities. 72+12=84, so you're correct. 

JP

0.28*53+7.50=22.34 dollars

JP

The first time she finished in 12 minutes. 

The second time she finished in 12/2=6 minutes

The third time she finished in 12/3=4 minutes

JP

Also, could you please help me out with part a? Is it 3^4?

Thank you so much! This really helped me out!

1520-950 = 570
570/950 = 0.6 = 60%

or is the answer 79.2 ?

cuz I assumed that rotating sprinkler rotates twice first so 60 degrees, not 30
which of these 2 answers is right? im confused
 

Are you happy with guest's answers?   (Perhaps you have a final answer sheet)

If you are then at least express some gratitude.

If not, then say so,  people are happy to interact with question askers that are genuinely trying to learn.

What is the sum of all values of y for which the expression (y + 6)/(y^2 - 3y - 2) is undefined?

The denominator cannot be 0

\(y^2-3y-2 \ne 0\\ y\ne\frac{3\pm\sqrt{9+8}}{2}\\ \)

So as JP said, the sum of these is 3

Find positive integers (a,b) so that sqrt(37 + 12*sqrt(7)) = a + b*sqrt(7).

\(\begin{array}{|rcll|} \hline \sqrt{37 + 12*\sqrt{7}} &=& \sqrt{9+28 + 12*\sqrt{7}} \\ &=& \sqrt{9+ 12*\sqrt{7} + 28 } \\ &=& \sqrt{9+ 2*6*\sqrt{7} + 28 } \\ &=& \sqrt{\left(3+2\sqrt{7}\right)^2 } \\ \mathbf{\sqrt{37 + 12*\sqrt{7}} } &=& \mathbf{3+2\sqrt{7} } \\ \hline \end{array}\)

Square root is an increasing function, so we only need to find an integer $n\ge100$ such that $\{\sqrt{n-1}\} \le 0.5$ but $\{\sqrt{n}\} >0.5$.  Since $\sqrt{110} = 10.488+$ and $\sqrt{111} = 10.535+$ so the smallest such $n$ is 111.

In a geometric sequence of real numbers,

the sum of the first 2 terms is 7,

and the sum of the first 6 terms is 91.

What is the sum of the first 4 terms?

\(\begin{array}{|rcll|} \hline s_2 &=& a_1 + a_2 \\ \mathbf{s_2} &=& \mathbf{a_1 + a_1r} \\ \hline \end{array} \begin{array}{|rcll|} \hline s_4 &=& a_1 + a_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_1r^2 + a_1r^3 \\ s_4 &=& s_2 + (a_1+a_1r)r^2 \\ s_4 &=& s_2 + s_2r^2 \\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\ \mathbf{s_4} &=& \mathbf{ 7(1+r^2) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s_6 &=& a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_1r^4 + a_1r^5 \\ s_6 &=& s_4 + (a_1+a_1r)r^4 \\ s_6 &=& s_4 + s_2r^4 \quad | \quad s_6 = 91,~s_2=7\\ 91 &=& s_4 +7r_4 \\ \mathbf{s_4} &=& \mathbf{ 91-7r_4 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s_4 = 7(1+r^2) &=& 91-7r_4 \\ 7(1+r^2) &=& 91-7r_4 \quad | \quad : 7 \\ 1+r^2 &=& 13-r^4 \\ \mathbf{r^4+r^2 -12} &=& \mathbf{ 0 } \\ \hline r^2 &=& \dfrac{ -1\pm\sqrt{1+4*12} }{2} \\\\ r^2 &=& \dfrac{ -1\pm 7 }{2} \\\\ r^2 &=& \dfrac{ -1 {\color{red}+} 7 }{2} \\\\ \mathbf{r^2} &=& \mathbf{3} \\ \hline s_4 &=& 7(1+r^2) \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}\)

Given x^2 + y^2 – 18x – 24y + 124 = 0  .

Find the radius: KEY POINTS to recognize

1) Know that the general equation of a circle is x² + y²  +2ax +2by +C = 0.

2) The center of the circle is given by (-a, -b).

3) The radius is given by  \(r= {\sqrt{a^2+b^2 -c} }\)

4) We can use radius formula above when

: the coeffitients of x²  and y²  are equal and no "xy"  term.

5 Then :  -18ax  means  (a = -18/2 =, -a =  -9)  and  -24by (-b = -24/2 = -12), C = 124.

Answer radius  \(r = {\sqrt{(-9)^2 +(-12)^2 - 124}}\)

\(r = {\sqrt{ 81+144 - 124}}\)

The greatest number is 55 McNuggets.