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Let P be a point on the graph of equation, (x-94)^2 + (y-4)^2 = 44^2. What is the shortest possible distance between P and the x-axis?

The centre of the circle is (94,4)  and the radius is 44

So the centre is in the first quadrant.

Let P be the point (x,y)

The domain is   [50,138]     The range is  [   ]

\((x-94)^2+(y-4)^2=44^2\\ (y-4)^2=44^2-(x-94)^2\\ If\;\;y\ge4 \;\;then\\ y=\sqrt{44^2-(x-94)^2}+4\\ \text{Distance to x axis is }\\ d=\sqrt{44^2-(x-94)^2}+4\\~\\ If\;\;y<4 \;\;then\\ y=-\sqrt{44^2-(x-94)^2}+4\\ \text{Distance to x axis is }\\ d=|-\sqrt{44^2-(x-94)^2}+4|\\~\\ \)

Here is an interactive graph.

Find the distance between \(P_1(3,-4)\) and the line 4x+8y+7=0.

Hello Guest!

\(f_1(x)= -\frac{1}{2}x-\frac{7}{8}\\ P_1(3,-4)\)

a. We find the line that goes through \(P_1\) and is perpendicular to the line \(y=-\frac{1}{2}x-\frac{7}{8}\) .

b. We calculate the distance from \(P_1\) to the intersection of the lines.

\(m_1=-0.5\\ m_2=-\frac{1}{m_1}=2\)

\(f_2(x)=m_2(x-x_1)+y_1=2(x-3)-4\\ \color{blue}f_2(x)=2x-10 \)

\(-\frac{1}{2}x-\frac{7}{8}=2x-10\\ 2.5x=\frac{73}{8}\\ x_2=3.65\\ y_2=2\cdot 3.65-10\\ y_2=-2.7\)

\(P_2(3.65,-2.7)\)

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

\(d=\sqrt{(-2.7-(-4))^2+(3.65-3)^2}\)

\(d=1.453\)

Te distance between \(P_1(3,-4)\) and the line 4x+8y+7=0 is 1.453.

  !

a/c   =   a/b * b/c

\(\frac{a}{c}\ =\ \frac{a}{b}\cdot\frac{b}{c}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt{10}}{\sqrt{21}}\cdot\frac{\sqrt{14}}{\sqrt{15}}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt{10\ \cdot\ 14}}{\sqrt{21\ \cdot\ 15}}\\~\\ \frac{a}{c}\ =\ \sqrt{\frac{10\ \cdot\ 14}{21\ \cdot\ 15}}\\~\\ \frac{a}{c}\ =\ \sqrt{\frac49}\\~\\ \frac{a}{c}\ =\ \frac{\sqrt4}{\sqrt9}\\~\\ \frac{a}{c}\ =\ \frac23\)

.   .

Since  (a, b)  is on the graph of  y = f(x),  we can say that:   f(a) = b

Let's plug in (a-6)/3  for  x  into the equation for the second graph.

y  =  -3 * f( 3 [ (a-6)/3 ] + 6)  -2   =   -3 * f(a) - 2   =   -3b - 2

When  x = (a-6)/3 ,  y = -3b - 2  so we can say that the point  \((\frac{a-6}{3}, -3b-2)\)  must be on the second graph.

2 nines    *99, 9*9,  99*     8+9+9 = 26

3 nines    999                               =  1

1 nine

**9             8*9=72

*9*             8*9 = 72

9**            9*9=81

so I get       26+1+72+72+81  = 

\(g(x)=(x^2-x+1)e^x\)

The function is continuous in the interval [-4, 1] so, to find its maximum value, we only have to examine the endpoints of the interval and the points at which the slope (derivative) is 0.

First let's find all the values of  x  that make the derivative equal  0 .

\(g'(x)=(x^2-x+1)(e^x)+(e^x)(2x-1)\)

Set the derivative equal to 0 and solve for  x :

\((x^2-x+1)(e^x)+(e^x)(2x-1)=0\\~\\ e^x(\ (x^2-x+1)+(2x-1)\ )=0\\~\\ e^x(\ x^2+x\ )=0\\~\\ e^x(x)(x+1)=0\)

Set each factor equal to zero and solve each equation for  x:

\(\begin{array}{} e^x=0&\qquad\text{or}\qquad &x=0&\qquad\text{or}\qquad& x+1=0\\~\\ \text{no solution}&&x=0&&x=-1 \end{array}\)

So only the following x values have the possibility to produce the maximum  g  in the interval:  -4, -1, 0 1

\(\begin{array}{} g(-4)&=& ((-4)^2-(-4)+1)e^{-4}&=& \frac{21}{e^4}& \approx&0.385\\~\\ g(-1)& =& ((-1)^2-(-1)+1)e^{-1}& =&\frac{3}{e^1}&\approx& 1.104\\~\\ g(0)&=& ((0)^2-(0)+1)e^{0}& =& (1)1& =& 1\\~\\ g(1)&=& ((1)^2-(1)+1)e^{1}& =& (1)e^1& \approx& 2.718 \end{array}\)

The maximum value of  g  in the interval  [-4, 1]  is  2.718  and it occurs when  x = 1

I'd like to see an answer for this too.    

\(3\lt\sqrt{2x}\lt4\\~\\ 3^2\lt(\sqrt{2x}\ )^2\lt4^2\\~\\ 9\lt2x\lt16\\~\\ \frac92\lt\frac{2x}{2}\lt\frac{16}{2}\\~\\ 4.5 \lt x \lt 8\)

The integers between 4.5 and 8 exclusively are:

5, 6, 7

So there are 3 different integer values of  x  that satisfy the condition.

Angle B = 15 degrees

5, 7, 9, 11     are consecutive odd integers.

I do not think you copied the question properly

1/2 of the x coefficient       squared     = 169       you will need to subtract this 

(x+13)^2  -169 = 33

x^2 +26x + 169  -169   = 33

x^2 + 26x =33    

The two dashed line segments intersect at an angle of 22 degrees.

Hello Guest,

I edit it first and my start solution was also 599.97, but I was a minimal bit unsure,

I'm not going to change my answer, I'd rather leave it, 

thank you.

Straight

Hello Guest,

\(\mathcal { \mbox {Is the expression on the right supposed to be (ab)²  or  a • b²   ? }}\)

In this case \(a^2+b^2=ab^2\) , whereby one can also rewrite this to \(a \cdot b^2\) .

Straight

Three consecutive positive odd integers????

OOps.... you listed the other two possibilities, but didn't count them   4 4   and   1 1

777,777,777,777^2 - 222,222,222,222^2

Total Sum = 108
Total Num = 24

 Find n.

Hello Guest!

\(x^2 + mx + (2m+n) = 0\\ x=-\frac{m}{2}\pm \sqrt{(\frac{m}{2})^2-2m-n}\\ n\le m( \frac{m}{4}-2)\)

if m = 8 then  n = 0

so

x = - 4

  !

6 x 6 = 36 possible rolls

two squares rolls include   1 1    1 4     4 1   and 4 4

4 out of 36      = 1/9

9 savory   *  5 C 2 sweet   = 9 x 10 = 90 plates

Evaluate the expression. Write your answer in scientific notation.
(7x10-3)+0.32-(6x10-2)

Hello Guest!

(7x10-3)+0.32-(6x10-2)

\(=\color{blue}7\cdot 10^{-3}+0.32-6\cdot10^{-2}\\ =0.007+0.32-0.06\\ =0.267\\ \color{blue} =2.67\cdot 10^{-1}\)

  !

Even though there are 2 dice, but they are tossed ONCE simultabeously as follows:

(1, 1) ,  (1, 2) ,  (1, 3) ,  (1, 4) ,  (1, 5) ,  (1, 6) ,  (2, 1) ,  (2, 2) ,  (2, 3) ,  (2, 4) ,  (2, 5) ,  (2, 6) ,  (3, 1) ,  (3, 2) ,  (3, 3) ,  (3, 4) ,  (3, 5) ,  (3, 6) ,  (4, 1) ,  (4, 2) ,  (4, 3) ,  (4, 4) ,  (4, 5) ,  (4, 6) ,  (5, 1) ,  (5, 2) ,  (5, 3) ,  (5, 4) ,  (5, 5) ,  (5, 6) ,  (6, 1) ,  (6, 2) ,  (6, 3) ,  (6, 4) ,  (6, 5) ,  (6, 6) ,  Total == 36

Therefore, there are only 2 possibilities of tossing 2 squares: (1, 4) and (4,1): The probability is 2 / 36 ==1/18