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 #3
avatar+23521 
+1

Solve for real numbers x, y, z:
x + yz = 6,
y + xz = 6,
z + xy = 6.

 

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{x + yz} &=& \mathbf{6} \\ & yz &=& 6-x \\ & \mathbf{z} &=& \mathbf{ \dfrac{6-x}{y} } \quad | \quad \boxed{ y\neq 0 !} \\ \hline (2) & \mathbf{y + xz} &=& \mathbf{6} \\ & xz &=& 6-y \\ & x\left(\dfrac{6-x}{y}\right) &=& 6-y \\ & x(6-x) &=& y(6-y) \\ & 6x-x^2 &=& 6y-y^2 \\ & \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ \hline (3) & \mathbf{z + xy} &=& \mathbf{6} \qquad \text{ or } \qquad \mathbf{z = 6-xy} \\ & xy &=& 6-z \\ & xy &=& 6-\left(\dfrac{6-x}{y}\right) \\ & xy &=& \dfrac{6y-(6-x)}{y} \\ & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ x &=& \dfrac{ 6\pm \sqrt{36-4(6y-y^2)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4y^2-24y+36} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y^2-6y+9)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y-3)^2} } {2} \\ x &=& \dfrac{ 6\pm 2(y-3) } {2} \\ x &=& 3 \pm (y-3) \\ \\ \mathbf{ x = y } \quad &\text{or}& \quad \mathbf{x=6-y} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline x_1 = y: & y^3 &=& 6y-6+y \\ & y^3-7y+6 &=& 0 \\ & y_1 &=& -3 \qquad \Rightarrow x_1 = y_1 = -3 \\ & y_2 &=& 1 \qquad \Rightarrow x_2 = y_2 = 1 \\ & y_3 &=& 2 \qquad \Rightarrow x_3 = y_3 = 2 \\ \hline x_2 = 6-y: & (6-y)y^2 &=& 6y-6+(6-y) \\ & 6y^2-y^3 &=& 5y \\ & y^3- 6y^2 +5y &=& 0 \\ & \underbrace{y}_{y\neq 0} \underbrace{(y^2- 6y +5)}_{=0} &=& 0 \\\\ & y^2- 6y +5 &=& 0 \\ & (y-5)(y-1) &=& 0 \\ & y_4 &=& 5 \qquad \Rightarrow x_4 = 6-y_4 = 1 \\ & y_5 &=& 1 \qquad \Rightarrow x_5 = 6-y_5 = 5 \\ \hline \end{array}\)

 

\(\begin{array}{|c|r|r|r|} \hline & x & y & z= 6-xy \\ \hline 1. & -3 & -3 & -3 \\ 2. & 1 & 1 & 5 \\ 3. & 2 & 2 & 2 \\ 4. & 1 & 5 & 1 \\ 5. & 5 & 1 & 1 \\ \hline \end{array}\)

 

laugh

1 hour ago
 #2
avatar+105935 
+1

 

Find the sum of all the two digit numbers which can be expressed as an integral multiple of the product of their digits.

 

I did this by looking at the numbers. It is not as arduous as it looks

 

10a+b=kab       where  1<= a <= 9       0<=b<=9     and k is a positive integer

 

If a=1 then

10+b=1b * k

b=0 no

b=1   k=11  yes   11

b=2    k=6  yes    12

b=3    no

b=4    14 is not a multiple of 4 so no

b=5     15 is a multple of 5  k= 3         15

b=6    16 is not a multiple of 6 so no

b=7     17 is not a multiple of 7 so no

b=8     18 is not a multiple of 8 so no

b=9     19 is not a multiple of 9 so no  

 

Ok now I can see a bit of a pattern emerging.  The original number must be a multiple of the last digit.

so when I look at a=2  I only need think about even 2 digit numbers

10a+b=kab 

If a=2 then

20+b= 2b * k

b=2  no

b=4    24 is divisable by 2*4=8  good    24

b=6  no, b=8 no

 

10a+b=kab    a=3

30+b=3b * k

If a=3 I only need consider numbers that are multiples of 3 which are  33,36,39

divide by 3  and I get   11,12,13

these must also be multiples of by b  

11/3  no,   12/6 yes  13/9 no

SO only   36 works here

 

10a+b=kab    a=4

40+b=4b * k

If a=4 I only need consider numbers that are multiples of 4 which are  44 and 48

divide by 4  and I get   11 and 12

these must also be multiples of by b  

11/4 no   12/8 no

SO none of these work

 

10a+b=kab    a=5

50+b=5b * k

only 55 is a multiple of 5

55/5=11   11 is not a multiple of 5 so no good

 

10a+b=kab    a=6

66 does not work

nor will 77 or 88 or 99

 

So the only ones that work are     11,  12,  15,  24  and 36.

 

Guest has already added them for you.

4 hours ago
 #1
avatar+23521 
+2

Saturn has a radius of about 9.0 earth radii, and a mass 95 times the Earth’s mass.

Estimate the gravitational field on the surface of Saturn compared to that on the Earth.

 

Newton's law of universal gravitation : \(F = G \dfrac{Mm}{r^2}\)
Newton's laws of motion : \(F=ma\)

 

\(\begin{array}{|rcll|} \hline F=ma&=& G \dfrac{Mm}{r^2} \\\\ \mathbf{a} &=& \mathbf{\dfrac{GM}{r^2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline a_{\text{Earth}} &=& \mathbf{\dfrac{GM_{\text{Earth}}}{R_{\text{Earth}}^2}} \\ a_{\text{Saturn}} &=& \mathbf{\dfrac{GM_{\text{Saturn}}}{R_{\text{Saturn}}^2}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{GM_{\text{Saturn}}}{R_{\text{Saturn}}^2}\above 1pt \dfrac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ M_{\text{Saturn}}R_{\text{Earth}}^2}{R_{\text{Saturn}}^2M_{\text{Earth}}} \quad | \quad R_{\text{Saturn}} = 9R_{\text{Earth}},\ M_{\text{Saturn}}= 95M_{\text{Earth}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ 95M_{\text{Earth}}R_{\text{Earth}}^2}{\left(9R_{\text{Earth}}\right)^2M_{\text{Earth}}} \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& \dfrac{ 95R_{\text{Earth}}^2}{81R_{\text{Earth}}^2 } \\\\ \dfrac{a_{\text{Saturn}}} {a_{\text{Earth}}} &=& 1.17283950617 \\ a_{\text{Saturn}} &=& 1.17283950617 *a_{\text{Earth}} \quad | \quad a_{\text{Earth}}= 9.8\ \dfrac{m}{s^2} \\ \mathbf{ a_{\text{Saturn}} } &\approx& \mathbf{ 11.5\ \dfrac{m}{s^2} } \\ \hline \end{array}\)

 

laugh

5 hours ago
 #5
avatar
0

I worked this out a different way and got a weird answer. Could you check my work to see if it is correct? 

 

In order to solve this problem we should first find the probability of making a ball into the bin on the first shot. It is not the probability of making one in 5 shots divided by 5 because you can make multiple shots out of 5. 

We can find the probability of making a ball into the bin on the first shot by making an equation. First, we need to find the chance we don't make a ball in the bin in 5 shots. That is \(1-\frac{211}{243}=\frac{32}{243}\) Next we can set p as P(making a ball into the bin on the first shot). This is the same probability of all the other shots because the probability is uniform across every shot. In terms of p, the chance of not making a ball in all of the shot is\( (1-p)^5\). Now we can make an equation. We get \((1-p)^5=\frac{32}{243}\). We can take the 5th root of both sides to get \((1-p)=\frac{2}{3} \)}. Solving for p we get \(p=1-\frac{2}{3}=\frac{1}{3}\).

Next, we can use complementary counting to find the probability of getting at least 2 shots in. `We can say P(0) is the probability of making no shots. P(1) is the probability of making one shot and so on. We can use this to make an equation. The probability of making at least 2 shots is 1-P(0)-P(1).

The probability of making 0 shots is \((\frac{2}{3})^6=\frac{64}{729}\)}. Next we have to find the probability of making 1 shot. The probability of that is \((\frac{2}{3})^5 \cdot \frac{1}{3}=\frac{32}{729}\). Now, we can find the probability of making at least 2 shots by plugging in the probabilities. We get \(1-\frac{64}{729}-\frac{32}{729}=\frac{633}{729}=\frac{211}{243}\). So, the probability of making two shots in 6 tries is \(\frac{211}{243}\)}.

9 hours ago
 #1
avatar
+1

Solve for x:
(9 x^2 - 16)^3 + (16 x^2 - 9)^3 = (25 x^2 - 25)^3

Expand out terms of the left hand side:
4825 x^6 - 10800 x^4 + 10800 x^2 - 4825 = (25 x^2 - 25)^3

Expand out terms of the right hand side:
4825 x^6 - 10800 x^4 + 10800 x^2 - 4825 = 15625 x^6 - 46875 x^4 + 46875 x^2 - 15625

Subtract 15625 x^6 - 46875 x^4 + 46875 x^2 - 15625 from both sides:
-10800 x^6 + 36075 x^4 - 36075 x^2 + 10800 = 0

The left hand side factors into a product with seven terms:
-75 (x - 1) (x + 1) (3 x - 4) (3 x + 4) (4 x - 3) (4 x + 3) = 0

Divide both sides by -75:
(x - 1) (x + 1) (3 x - 4) (3 x + 4) (4 x - 3) (4 x + 3) = 0

Split into six equations:
x - 1 = 0 or x + 1 = 0 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 1 to both sides:
x = 1 or x + 1 = 0 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Subtract 1 from both sides:
x = 1 or x = -1 or 3 x - 4 = 0 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 4 to both sides:
x = 1 or x = -1 or 3 x = 4 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Divide both sides by 3:
x = 1 or x = -1 or x = 4/3 or 3 x + 4 = 0 or 4 x - 3 = 0 or 4 x + 3 = 0

Subtract 4 from both sides:
x = 1 or x = -1 or x = 4/3 or 3 x = -4 or 4 x - 3 = 0 or 4 x + 3 = 0

Divide both sides by 3:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or 4 x - 3 = 0 or 4 x + 3 = 0

Add 3 to both sides:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or 4 x = 3 or 4 x + 3 = 0

Divide both sides by 4:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or x = 3/4 or 4 x + 3 = 0

Subtract 3 from both sides:
x = 1 or x = -1 or x = 4/3 or x = -4/3 or x = 3/4 or 4 x = -3

Divide both sides by 4:
 
x = 1   or    x = -1    or    x = 4/3    or    x = -4/3    or    x = 3/4    or    x = -3/4

11 hours ago

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