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Thank you so much for your help!

Hello, Guest!

I am glad you attempted this problem. However, I cannot follow your line of reasoning, and your answer differs from mine. Could you explain your thought process in more detail so that people can follow your thought process?

In order to ease solving process, I have created a diagram that represents the situation to scale. The diagram is below with points I have labeled arbitrarily where O is the center of the circle.

Since O is the center of the circle, OC = OJ = r, the radius of the circle. In addition, we know that \(\overline{\rm BH}\) is a perpendicular bisector of the chords, so \(\rm BC = 9\) and \(\rm HJ = 7\). Also, we can label \(\rm BO = BO\) and \(\rm OH = 10 - BO\). Putting all this information together enables us to find the radius of the circle.

\(r^2 = 9^2 + {\rm BO}^2 \\ r^2 = 7^2 + \left(10 - {\rm BO}\right)^2 \\ 81 + {\rm BO}^2 = 49 + 100 - 20{\rm BO} + {\rm BO}^2 \\ 20{\rm BO} = 68 \\  {\rm BO} = \frac{17}{5} \\  r^2 = 81 + \left(\frac{17}{5}\right)^2 = \frac{2025}{25} + \frac{289}{25} = \frac{2314}{25}\)

We know that BO = 5, so \({\rm OE} = 5 - {\rm BO} = 5 - \frac{17}{5} = \frac{8}{5}\). We can apply Pythagorean's Theorem one more time to find DF, the length of the desired chord.

\(r^2 = {\rm OE}^2 + {\rm EF}^2 \\ \frac{2314}{25} = \left(\frac{8}{5}\right)^2 + {\rm EF}^2 \\ {\rm EF}^2 = \frac{2314}{25} - \frac{64}{25} \\ {\rm EF}^2 = 90 \\  {\rm EF} = 3\sqrt{10} \\ {\rm DF} = 2{\rm EF} = 6 \sqrt{10} \approx 18.9737\)

When combining three bags of marbles with 36%, 32%, and 25% yellow marbles respectively, the resulting proportion of yellow marbles in the bowl would be the weighted average of the percentages:

weighted average = (percentage of bag 1 * weight of bag 1 + percentage of bag 2 * weight of bag 2 + percentage of bag 3 * weight of bag 3)/(total weight)

Assuming each bag has the same weight (equal weight), the calculation becomes:

Weighted average = (0.36 + 0.32 + 0.25) / 3 ≈ 0.31 or 31%

So, approximately 31% of the marbles in the bowl would be yellow.

OP = x

OB = OD = OF = R

PB = 9^v RF = 7 PRD = 10

R^v = 9^v + x^v and R^v = 7^v + (10 - x)^v

both R equal 9^v + x^v = 7^v + 10^v - 20x + x^v

                    x = (7^v + 10^v - 9^v)/20

                       x = 68/20

                       x = 3.4

                      OP = 3.4, O9 = 5 - 3.4 = 1.6

                                       R = \({\sqrt{9^v-x^v}}\)

                                           = \({\sqrt{9^v-(3.4)^v}}\)

                                           = \({\sqrt{92.56}}\)

9D = \({\sqrt{(92.56^v-(1.6)^v}}\)

     = \({\sqrt{92.56^v-2.56}}\)

    = \({\sqrt{90}}\)

CD = 2\({\sqrt{90}}\)

\({Now\ Checking \over}\)
 

                       OR = OQ + QR

                              = 1.6 + 5 = 6.6

                 OF = R = \({\sqrt{92.56}}\)

           RF = \({\sqrt{92.56 - 43.56}} = {\sqrt{49}} = 7\)

So, own Solⁿ is correct:

The length of the chord that lies halfway between them is 2√90  units. 

n ==11,  167

[11 + 15]^2 mod [11 + 2] ==0

[167  +  15]^2 mod [167 + 2]==0

Sum of  all n:  11 +  167 ==178

Let's denote the coordinates of point $C$ as $(x, y)$. We are given that point $C$ is six times as far from point $A$ as it is from point $B$. This can be expressed as an equation:

Distance from $C$ to $A$ = 6 * Distance from $C$ to $B$

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian plane can be calculated using the distance formula:

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

In this case, we have:

Distance from $C$ to $A$ = $\sqrt{(x - (-1))^2 + (y - 0)^2}$ = $\sqrt{(x + 1)^2 + y^2}$

Distance from $C$ to $B$ = $\sqrt{(x - 3)^2 + (y - 8)^2}$

Now, we can write the equation based on the information given:

$\sqrt{(x + 1)^2 + y^2} = 6 \cdot \sqrt{(x - 3)^2 + (y - 8)^2}$

Squaring both sides of the equation to eliminate the square root:

$(x + 1)^2 + y^2 = 36 \cdot ((x - 3)^2 + (y - 8)^2)$

Expand the squared terms:

$x^2 + 2x + 1 + y^2 = 36 \cdot (x^2 - 6x + 9 + y^2 - 16y + 64)$

Now, let's simplify the equation:

$x^2 + 2x + 1 + y^2 = 36x^2 - 216x + 324 + 36y^2 - 576y + 2304$

Combine like terms:

$35x^2 - 218x + 2303 + 35y^2 - 576y = 0$

Divide the entire equation by 35 to simplify:

$x^2 - \frac{218}{35}x + \frac{2303}{35} + y^2 - \frac{576}{35}y = 0$

Now, we can complete the square for both the $x$ and $y$ terms to express the equation in the standard form of a circle:

$x^2 - \frac{218}{35}x + \left(\frac{109}{35}\right)^2 + y^2 - \frac{576}{35}y + \left(\frac{288}{35}\right)^2 = \left(\frac{109}{35}\right)^2 + \left(\frac{288}{35}\right)^2$

Factor the squared terms:

$\left(x - \frac{109}{35}\right)^2 + \left(y - \frac{288}{35}\right)^2 = \left(\frac{109}{35}\right)^2 + \left(\frac{288}{35}\right)^2$

Now, the equation is in the standard form of a circle:

$(x - \frac{109}{35})^2 + (y - \frac{288}{35})^2 = \left(\frac{109}{35}\right)^2 + \left(\frac{288}{35}\right)^2$

Comparing this to the standard form equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$, we can see that the center of the circle is $\left(\frac{109}{35}, \frac{288}{35}\right)$ and the radius is $\sqrt{\left(\frac{109}{35}\right)^2 + \left(\frac{288}{35}\right)^2}$.

So, the coordinates of point $C$ are approximately $\left(\frac{109}{35}, \frac{288}{35}\right)$.

Number of 4-digit palindromes that are divisible by 8:

(2112, 2552, 2992, 4224, 4664, 6336, 6776, 8008, 8448, 8888) >Total = 10 such palindromes.

Is it okay to just see two squares, side by side?  If so, the total area is 36 + 36 = 72 square units

_.

To solve this, we can set up a proportion equation:

x / 4 = 9 / y

Cross-multiplying gives:

xy = 36

Now, we need to find the pairs of positive integers (x, y) that satisfy this equation. The pairs that work are:

(1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (9, 4), (12, 3), (18, 2), (36, 1)

So, there are 9 pairs of positive integers (x, y) that satisfy the given property.

Therefore, the required pair of positive integers (x,y) satisfied the given property is 9.

We have the number $256

The proper divisors of $256 are $1$, $2$, $4, $8$, $16$, $32$, $64$, and $128$

Now 1 + 2 + 4 + ..... + 128; Here first term = 1 

                                                     common ratio = 2

       = \( {1 (2^8 - 1) \over 2 - 1}\)         [ a + ar + ar² + ... + ar^n-1

                                  = \({a(r^n - 1) \over r - 1}\);(r >1)

       = \( {256 - 1 \over 1}\)

       = 255

Thus the sum of proper divisor of $256$ is $255$

Nice job!

Assuming that \(O\) is the circumcenter of \(\triangle ABC\), the measure of x is \(128^{\circ}\)

I suppose you mean compound interest. There are 30 bi-months in five years. We can turn this is 24000*1.01^30, which equals to 32348.37, when rounded to the nearest hundreth. 

This will have two distinct roots when the  discriminant  >  0

So

6^2  - 4 (1)(c) > 0

36 - 4c > 0

36 > 4c

36/4  > c

9 > c

It will have two  distinct roots when    c < 9

I'm unsure if S is supposed to lie outside or inside of P.....here, I've assumed outside

Point FC = 1.....so  AF / sin 70° = FC / sin 20° so that   AF = (sin 70°)/(sin20°)

F = (0, sin 70°/sin 20°)

DF = 1

So....the radius of  S =   (1 + sin 70°/sin20°)

The exact perimeter =   2 ( 1 + sin70°/sin20°) pi  =  ( 2 + 2sin 70°/sin20°) pi ≈  7.5 pi