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sin x = (2 - sqrt(3))/2

Let the price of one unit of Brand Y soda be $y$, so the price of one unit of Brand X soda is $0.9y$.

According to the advertisement, the quantity of soda in one unit of Brand X soda is 20% more than the quantity of soda in one unit of Brand Y soda. Therefore, the ratio of the quantities of soda in one unit of Brand X and Brand Y soda is:

(1 + 0.2)/1 = 1.

So, one unit of Brand X soda has 1.2 times as much soda as one unit of Brand Y soda.

Therefore, the ratio of the unit price of Brand X soda to the unit price of Brand Y soda is:

0.9y/(y/1.2) = 0.9*y*1.2/y = 

$\frac{0.9y}{y/1.2} = \frac{0.9y \cdot 1.2}{y} = 27/25

b)

Suppose the coefficients are (n C r−1), (n C r), (n C r+1).

Then: [n - r + 1] / r ==8 / 1
[n - r] / [r +1] ==40 / 8, solve for n, r
n ==17  and  r ==2

So, the coefficients are: 17 C 1 : 17 C 2 : 17 C 3==
17 : 136 : 680 ==1 : 8 : 40

(a) \(\frac{\binom{n}{k}}{\binom{n}{k - 1}} = \frac{\frac{n!}{k! (n - k)!}}{\frac{n!}{(k - 1)! (n - k + 1)!}} = \frac{n!(k - 1)! (n - k + 1)!}{k! (n - k)! n!} = \frac{(k - 1)(n - k + 1)}{k}\)

(b) 

Let the expansion of (1 + x)^n be:

(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n)x^n

We know that:

a:b:c = 1:8:40

Therefore:

C(n, 0) : C(n, 1) : C(n, 2) = 1 : 8 : 40

Using the formula for the binomial coefficients, we have:

C(n, 0) = 1 C(n, 1) = n

Therefore:

C(n, 2) = 28C(n, 1) - 27C(n, 0)

Using the values we have:

C(n, 2) = 28n - 27

We know that:

C(n, 0) + C(n, 1) + C(n, 2) = 1 + 8 + 40 = 49

Therefore:

1 + n + C(n, 2) = 49

Solving this equation, we get n = 9.

1 quart = 4 cups

1 2/3 quarts x  4 ==6 2/3 cups

(6 2/3) / (2/3) =10 - glasses of juice that can be poured from the pitcher 

What is the smallest distance?

Hello Guest!

\(y = 0.5\cdot (x^2 - 18)\\ y=0.5x^2-9\\ r^2(x)=x^2+y^2\\ r^2(x)=x^2+(0.5x^2-9)^2\\ \frac{dr^2(x)}{dx}=2x+2x(0.5x^2-9)\\ \frac{dr^2(x)}{dx}=2x+x^3-18x\)

\( \frac{dr^2(x)}{dx}=x^3-16x=x(x^2-16)=0\\ x\in \{-4,0,{\color{blue}4}\}\\ y=0.5\cdot 16-9\\ \color{blue}y=-1\)

\(l=\sqrt{x^2+y^2}=\sqrt{16+1}\\ \color{blue}l=4.123\)

The smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 18) is 4.123 .

  !

X= -6

10cm

Hello,

The volume of the cylinder is approximately 12.57 cubic inches. This is calculated by using the formula V=πr^2h, where r is half the diameter (2 inches) and h is the height (1 inch).

Sorry for the bad wording. 
I meant to write the largest value of sin(x).

To get a constant term, we need the z's to cancel out. Notice that we can't eliminate the z with less than 4 \(2z\)'s and \({1 \over z \sqrt z}\)'s. The closest we can do is \((2z)^3 \times ({1 \over z \sqrt z})^2\), but that uses 5, so it doesn't work. 

Thus, the only constant term will be the 1 multiplied by itself 4 times, which is \(1^4 = \color{brown}\boxed{1}\)

Area of circumcircle==Pi * [a / sqrt(3)]^2  

Area of incircle ==Pi * [a /(2*sqrt(3))]^2

x  -  y ==Pi * [a / sqrt(3)]^2  -  Pi * [a /(2*sqrt(3))]^2

x  -  y==[Pi * a^2] / 4

Let \(x = \log_b ({2 \over 3})\) and \(y = \log_b ({2 \over 3})\)

Then \(b^x = {2 \over 3}\) and \(b^y = {3 \over 2}\)

Multiplying the two gives us \(b^x \times b^y = b^{x + y} = 1\)

Now, notice that the only way the expression equals 1 is if \(x + y = \color{brown}\boxed{0}\).

Note: In general, \(\log_b({x}) + \log_b({y}) = \log_b({xy})\)

Thank you!

You can get 14 glasses of juice.

After Tasha moves A to A′, A′ is 2​ units to the right of A and 3​ units above O. Therefore, A′=(sqrt2​,sqrt3​).

After Richard moves A′ to A′′, A′′ is 90∘ clockwise from A′ and the same distance from O. Therefore, A′′=(−sqrt3​,sqrt2​).

After Richard moves A to A1​, A1​ is 90∘ clockwise from A and the same distance from O. Therefore, A1​=(−sqrt3​,sqrt3​).

After Tasha moves A1​ to A2​, A2​ is 2​ units to the right of A1​ and 3​ units above O. Therefore, A2​=(−sqrt3​+sqrst2​,sqrt3​+sqrt2​).

Therefore, A′′A2​=(−sqrt3​+sqrt2​−(−sqrt3​))(−sqrt3​+sqrt2​−sqrt3​+sqrt2​) = sqrt(2)​.

At the veterinarian's office, Terri learned that her dog weighed 4 times as much as her cat. together the pets weighed 40 lbs. how much did the dog weigh.   

                                                   D  =  4C  

                                            D + C  =  40  

                                            4C + C  =  40  

                                                  5C  =  40  

                                                    C  =  8  

                                                    D  =  4C  

                                                    D  =  32  

_.

\(\log_{b} \left(\dfrac 2 3\right) + \log_{b} \left(\dfrac 3 2\right) = 1\)

\(\sqrt{7! + 6!} \)

\(\sqrt{6!(7 + 1)}\)

\(\sqrt{720 \times 8}\)

\(\sqrt{2^7 \times 3^2 \times 5}\)

\(24 \sqrt{2 \times 5}\)

\(\color{brown}\boxed{24 \sqrt{10}}\)

\((2x+10)^4(x+5)^4\)

\((2(x+5))^4(x+5)^4\)

\(16(x+5)^4(x+5)^4\)

\(\color{brown}\boxed{16(x+5)^8}\)

3+2/3x=3/4

Multiply every term by 12

36  +  8x = 9

8x = 9 - 36

8x =  - 27

Simplify the left-hand side to \(x^2 + 28x - 21\).

Now, notice that the right-hand side is just vertex form. This means that a is the coefficient of the \(x^2\) term (in this case it's 1) and that \((h, k)\) is the vertex. 

The vertex is located at \(x = -{b \over 2a} = -14\). Substituting this in gives us \((-14)^2 + 28(-14) - 21 = -217\). So, the vertex is \((-14, -217)\).

Thus, the ordered triple is just \(\color{brown}\boxed{(1, -14, -217)}\)

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