BuilderBoi

By the quadratic formula, the roots are: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).

Substituting what we know, we have: \(x = {-99 \pm \sqrt{99^2-2c}}\)

Now, we have the equation: \(99^2 - 2c = 9959\)

Can you take it from here?

By the property of geometric sequences, \(g_{43} = g_1 \times d^{42}\). 

Note that \(g_{28} = g_{23} \times d^5\). Substituting the given values, we have: \(28 = 16d^5\), meaning \(d = \sqrt[5]{7 \over 4}\).

Now, notice that \(g^{43} = g^{28} \times d^{15}\). 

We know that \(\sqrt[5]{7 \over 4} = {7 \over 4}^3 = {343 \over 64}\)

This means that \(d^{43} = 28 \times {343 \over 64} = \color{brown}\boxed{150 {1 \over 16}}\)

Because both equations are equal to y, we can set them equal to each other: \(3x^2 - 6x = 2x^2 + x - c\)

Bringing everything to the right-hand side gives us the quadratic: \(x^2 - 7x + c = 0\)

For the quadratic to have exactly 1 solution, the discriminant (\(b^2 - 4ac\)) must equal 0. This means \(49 - 4c = 0\)

For the quadratic to have more than 1 solution, the discriminant must be greater than 0. This means: \(49 - 4c >0\)

For the quadratic to have no real solutions, the discriminant must be less than 0. This means:\(49 - 4c <0\)

Can you take it from here?

Moving everything to the right-hand side gives us: \(x^2 - 3x + 2 = 0 \)

Let a and b be the roots of the quadraitc.

Now, recall the identity: \(a^2 + b^2 = (a+b)^2 - 2ab\)

By Vieta's, we know that \(a + b = -{ b\over a} = 3\) and \(ab = {c \over a} = 2\).

Can you take it from here?

the answer is 

4765444444444444748563785678563756347563478563474364784684765483654+487475643756475657r5643565-86767556875436+65497856437856437564378564375867854368753647854656435784545643573604416504576475405314065465416000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000033648564865329645327657645364354364532654532643564-374456854876354635467+4875174654847547643547454656764345678901234568976345678854678746734567871234567813456788123456789543345678833456775433x5456744678343256783754576437856237546547653478564875634785656576435766345764543465347653654+758643756437856475678568756445874565874-55737572565873676+587457643473654354655555555555559378565184764638546525475e456784323456775432345678732345432123456094234587654323456789876543

The \over function is used for fractions, so example, 1/2 is 1 \over 2

Curly brackets ({), are the equivalent of parenthesis in PEMDAS, so, for example, 1/2 = 0.5, would be {1 \over 2} = 0.5. The brackets around the fraction separate the fractions, without them, it would be \(1 \over 2 = 0.5\). 

Next, the square root function is \sqrt{} (where the number is inside the bracket), so, \(\sqrt {209}\) is \sqrt{209}.

So, \({17 - \sqrt {209}} \over 4\) is {17 - \sqrt{209}} \over 4.

To box something, use the \boxed{} function, where the thing that is being boxed goes inside the curly brackets. 

To make it a different color, use the /color{insert color here}{}, where the color goes in the first pair of brackets and the thing that you would like to change he color of. 

Putting all these concepts together, \(\color{brown}\boxed{17 - \sqrt {209} \over 4}\) in raw form is "\color{brown}\boxed{17 - \sqrt {209} \over 4}". 

/color{brown} makes everything brown, /boxed boxes the entire thing in brown, and the brackets immediately after that define the fraction that is being boxed. The \sqrt{209} makes the square root of 209, and \over 4 makes a fraction with a denominator of 4 and a numerator with everything before it ( in this case 17 - \sqrt {209}).

More info available here: https://web2.0calc.com/questions/latex-coding

Also, sorry if this isn't helpful, I'm not the best at explaining...

We can set 30 equal to y, and we have the equation: \(30 = -6t^2 + 51t\)

Subtracting 30 from both sides, we get: \(0 = -6t^2 + 51t - 30\)

Now, we can use the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), to get \({17 \pm \sqrt {209}} \over 4\).

But, we are looking for the first time it reaches 30 ft, so the answer is \(\color{brown}\boxed{{17 - \sqrt {209}} \over 4}\)

Note that 5 and 8 both multiply to 40. 

This means that \(3x^2 = 0\) and \(2y^2 = 0 \)

Can you take it from here?

One of the vertices is 6 units to the right of the vertex, so the other must be 6 to the left of the vertex. 

Can you find it from here?

https://web2.0calc.com/questions/math-problem_32546