1.25p=15, then we divide 1.25 on both sides to get 12 then our answer would be 12. p=12
Horrible x 1,000
none(if y=0) since we use the quadratic equation and we use a calculator we got our answer :D
thanks for the help guys
sinA = 2(cosA)
sin(A) / cos(A) = tan(A)
tan(A) = 2 ∠A = 63.43494882º
cos(A) = 0.447213595
Okay thank you so much for your help! I will work from here.
angle CDA=angle CEB corresponding angles on parrallel lines
If angle CAD = angle CBE what would have to be true about angle ACD and angle BCE?
I will let you finish it.
the solution is: -9/2 < t ≤ 9/4
so the interval is: \((-\frac{9}{2},\:\frac{9}{4}]\)
Thanks!
it simplifies to \(10x^2-16x+51\)
Draw a venn diagram and you can find your answer
Hi there! Welcome! I am pretty new here too, but here is what I know. When you go to the home page, there is an advanced calculator which can help you solve equations. In the forum (this is the forum and you just posted in it) you can help others solve problems and ask questions yourself. That's pretty much all I know. Hope this helped
Not that hard,
\(a=2-9a\)
\(=10a=2\)
\(=a=\frac{1}{5}\)
C
The product of my numbers is 40.
Lisa earns 45 dollars after five years.
Help?
One of the trig identities state that \(\tan A = \frac{\sin A}{\cos A}\) (you could just prove it by using sohcahtoa), so that's what we need to solve for. To do that, divide both sides by cos A and then by 2 to get:
\(\frac{\sin A}{\cos A} = \boxed{\frac{3}{2}}\), which is our final answer.
Note that the track is just a split up circle nad rectangle.
For the rectangle part, the outside is 160. (80*2)
For the circle part, the circumfrence is 35pi. (diameter = 35)
So running aorund the track is 160 + 35pi
=^._.^=
The solution is (-3/7, 10/3]
24,511 N would be the answer
Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 - 8x + 1. Find u/v + v/u.
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} \\ &=& \dfrac{u^2+v^2}{uv} \\\\ &=& \dfrac{u^2+v^2}{uv} +2-2 \\\\ &=& \dfrac{u^2+v^2}{uv} +\dfrac{2uv}{uv}-2 \\\\ &=& \dfrac{u^2+2uv+v^2}{uv}-2 \\\\ &=& \dfrac{(u+v)^2}{uv}-2 \\\\ \mathbf{\dfrac{u}{v}+\dfrac{v}{u}}&=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \\ \hline \mathbf{x^2+13x+6}&=&\mathbf{0} \\ \text{By Vieta: } \qquad u+v = -13, \quad uv = 6 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{u}{v}+\dfrac{v}{u}} &=& \mathbf{\dfrac{(u+v)^2-2uv}{uv}} \quad | \quad u+v = -13, \quad uv = 6 \\\\ &=& \dfrac{(-13)^2-2*6}{6} \\\\ &=& \dfrac{169-12}{6} \\\\ &=& \mathbf{\dfrac{157}{6}} \\ \hline \end{array}\)
thank you but i can't click the heart :|
From SOHCAHTOA, we conclude that sin A = BC/AC and cos A = AB/AC. We have BC/AC=2AB/AC, or BC=2AB. You want tan A, which is BC/AB by SOHCAHTOA. But you have BC=2AB, so BC/AB=2, which is your answer.
