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Try visualizing this by drawing a graph labelled with the heights. :))

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I think some of these are people cheating on homework assignments or tests. 

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For quadratic \(ax^2+bx-c\), \(a < 0\). Its graph is a negative parabola. Let's calculate when the cannonball will fly exactly 8.7 meters. Substitute t into \(h(t) = -4.9t^2+14t-0.4\) which \(h(t)\) is 8.7 meters.  

Assume \(x = t\)

\(-4.9x^2+14x-0.4=8.7\)

\(-4.9x^2+14x-9.1=0\)

\(4.9x^2-14x+9.1=0\)

Substitute into the quadratic formula\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).

We get\(x = {-(-14) \pm \sqrt{(-14)^2-4*4.9*9.1} \over 2*4.9}\)

\(x = {14 \pm \sqrt{196-178.36} \over 9.8}\)

\(x = {14 \pm \sqrt{17.64} \over 9.8}\)

\(x = {14 \pm 4.2 \over 9.8}\)

\({x}_{1} = \frac{13}{7}\) or \({x}_{2} = 1\)

the region is \(\left |{x}_{2} - {x}_{1} \right |\), which equals \(\frac{6}{7}\)

2016/6 = 336

For each of those groups of 6, we can write it as 2 2 2 or 3 3. 

We can choose 0 "2 2 2" all the way up to 336 "2 2 2". 

336 + 1 = 337

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\(a^2-b^2+2a\)

\(=(a+b)(a-b)+2a\)

\(=-1(a-b)+2a\)

\(=-a-(-b)+2a\)

\(=b-a+2a\)

\(=b+a\)

\(=-1\)

It is \(\frac{201}{3}\), which is 67 mi./h

Dana has only nickels in her hand,

and Ajah has exactly the same number of dimes and no other coins.

Together they have a total of 90¢. How many coins is each person holding?

1. N = D

2. N*5+D*10 = 90

    N*5+N*10 = 90

    15N = 90

    N = 6 or D = 6

We have $y = \frac{x - 3}{x + 1} = 1 - \frac{4}{x+1}$.  So $\frac{4}{x+1} = 1-y$.  Since $x\ne -1$ we can multiply $x+1$ to both sides to get $(x+1)(1-y)=4$, or equivalently, $(x+1)(y-1)=-4$, i.e., $a=1$, $b=-1$ and $c=-4$.

Substituting $b=6$ into $ab-6a+5b=15$ we get the absurdity $30=15$.   Thus $b$ can never assume the value 6.   By algebraic manipulation we can express $a$ in terms of $b$ as $a = (15-5b)/(b-6)$.    Now, $15-5b > 0$ iff $b < 3$ and $b-6 > 0$ iff $b > 6$.   Therefore, $a$ is positive only when $3 < b < 6$.   Among these two plausible values of $b$ (4 and 5), only $b=5$ turns $a$ into a positive integer $a=10$.  Thus, $|a-b|$ has only one value 5 and it must be the minimum.

Let radius be r inches. 

Let's assume the corner of the wall is at (0,0) and P be the point, then the centre of circle is at (r,r).

Then equation of  circle is

      \((x-r)^2+(y-r)^2= r^2\)

And, according to question, points (18,35) and/or (35,18) lies on the circle.

Take any of them it doesn't make any difference

⇒ \((18-r)^2+(35-r)^2=r^2\)

⇒ \(324-36r+r^2+1225-70r+r^2=r^2\)

⇒ \(r^2-106r+1549=0\)

Solving this we get 

\(r=88.5 \)  and  \(r=17.5\)

Hence, the radius can be either 88.5 inches or 17.5 inches.

It seems there's not enough information to find $f(0)+f(4)$;  I need to know the value of $f(x)$ at one more point.

It turns out that the second constraint alone completely determines the values of $f(-1)$, $f(0)$, and $f(1)$.   The first constraint just makes it possible to determine the values of $f$ at other domain points.

We set $x = 0, 1, -1$ in order in the second constraint to get these systems of equations
\begin{eqnarray*}
    f(-1) &=& f(0)^2-2 \\
    f(0) &=& (f(1)-1)^2+1-2 \\
    f(0) &=& (f(-1)+1)^2+1-2
\end{eqnarray*}
Solving we find 3 solutions $(f(-1), f(0), f(1)) = (-1,-1,1)$ or $(-2,0,0)$ or $(-2,0,2)$.
The answer you are looking for is any one of $(-1,1)$, $(0,0)$, or $(0,2)$.

It turns out that the second constraint alone completely determines the values of $f(-1)$, $f(0)$, and $f(1)$.   The first constraint just makes it possible to determine the values of $f$ at other domain points.

We set $x = 0, 1, -1$ in order in the second constraint to get these systems of equations
\begin{eqnarray*}
    f(-1) &=& f(0)^2-2 \\
    f(0) &=& (f(1)-1)^2+1-2 \\
    f(0) &=& (f(-1)+1)^2+1-2
\end{eqnarray*}
Solving we find 3 solutions $(f(-1), f(0), f(1)) = (-1,-1,1)$ or $(-2,0,0)$ or $(-2,0,2)$.
The answer you are looking for is any one of $(-1,1)$, $(0,0)$, or $(0,2)$.

No that's absolutely wrong! Idk who you're but pls stop giving wrong answers on every question you find. You're not obliged to answer, if you don't know just leave it. 

I think the question contains a typo: $f^{-1}(x)$ should have the form $(ax+b)/(cx+d)$ instead.

Equation of vertical parabola is 

              (x - h)² = 4a(y - k)

where (h,k) is the vertex. 

⇒ (x - 2)² = 4a(y - 4)

∵ It passes through the point (1,17)

(1 - 2)² = 4a(17- 4)

52a = 1

a = 1/52 

∴ Equation of parabola is 

(x - 2)² = 4/52 (y - 4)

13(x² - 4x + 4) = y - 4

y - 4 = 13x² - 52x + 52

y = 13x² - 52x + 56

It's vertex is at (2,4) and it passes through (1,17). 

Since $70 = 2\cdot 5\cdot 7$, there are 8 ways to factor 70 into two factors (not taking order into consideration): $1\cdot 70$, $2\cdot 35$, $5\cdot 14$, $7\cdot 10$, and four other ways similar to these but where both factors are negative.  By checking all possibilities under the constraints $3b+c=41$ and $bc=70$, we find that only $b=2$ and $c=35$ satisfy both constraints.    Substituting $b=2$ into the first constraint $2a+3b=52$ gives $a=23$.  So $a+b+c = 23+2+35 = 60$.

p(x) = x²

q(x) = 2x³ + 1

r(x) = 7x⁴ - 5x³ - 6x - 2

p(q(x)) = p(2x³ + 1)  = (2x³ + 1)²  =  4x⁶ + 4x³ + 1

⇒ p(q(x)) + r(x) = 4x⁶ + 4x³ + 1 + 7x⁴ - 5x³ - 6x - 2

                         = 4x⁶ + 7x⁴ - x³ - 6x - 1

The formula of CI is 

                \(A=P(1+ {r\over n})^{nt}\)

where A = final amount

           P = principal amount

           r = rate

           t = time (in years) 

           n = no. of times compounded

So, here P = 10000, r = 10% = 0.1, n = 2, t = 1 year. 

Now, simply put the values

\(A= 10000(1+ {0.1\over 2})^2\)

\(A=$11,025\)

~Hope you got it. 

a/c = 4.

p(q(x)) + r(x) = 11x^4 - 8x^3 + 5x^2 - 6x + 7.

Hi Amy,

The question only asks for the arc (an angular measure) of LPM, in degrees. It gives the arc of LM as 72 degrees. 

All circles have 360 degrees, and all angular measurements of a circle are constant for any radius.

Of course, the linear measure of the arc length is proportional to the radius. 

--. .-

GA

Using similar triangles, WJ/WK = 28.