1) whenever you see log, think 3 (the base) to what power equals 1000 (the value in parentheses). 3^x = 1000. Unfortunately 3 doesn't go into 1000 so you're gonna get a decimal answer. The problem is that base 3... We can use the base formula to restate it... Log10(1000) / log10(3) = log3(1000)
10^x = 1000 is easy, that's just 3. So 3/log10(3). You can probably put that in your calculator since most have a log function
Replace x by \(x=r\cos \theta\) and y by \(y=r\sin \theta\)
The general equation of a circle can be written as: \((x-x_c)^2+(y-y_c)^2=r^2\) where the centre has coordinates \((x_c,y_c)\) and the circle has radius \(r\). We can find \(r\) from \(2\pi r = circumference\)
This should be enough for you to answer the question.
A bag of cat treats costs $5.98. Santiago bought 6 bags of cat treats. How much did Santiago spend in all?
Or in english .No, he will still need to run 50 more meters.
This is almost identical to the question I just answered so you can do it yourself.
We have,
\(sin(x+{\pi \over 2})-cos(x-\pi)=1\)
⇒\(cosx-cos(x-\pi)=1\)
Now, we know the formula \(cosx-cosy=-2sin{x+y\over 2}sin{x-y \over 2}\)
⇒\(-2sin{x+x-\pi\over 2}sin{x-x+\pi\over 2}=1\)
⇒\(-2sin({x-{\pi\over 2}})sin{\pi\over 2}=1\)
⇒\(2sin({\pi\over 2}-x)sin{\pi\over 2}=1\)
⇒\(2cosxsin{\pi\over 2}=1\)
⇒\(cosx=1/2\)
⇒\(x={\pi\over 3}\)
~Hope this helps :)
81.86%
^3√2x + 9^3√2x - 2^3√2x =
Is 2x under the cube roo, of just the 2? You should have used brackets.
Draw the triangle and use Pythagoras's theorem
Yeah you're right, in that case lets eliminate one possibility and make it 494 possible polynomials.
My bad! Thank you for correcting me and providing reference graphs, it's very appreciated. :)
No problemo :D
EZ the best answer ever THANK U
Plug in the values: \(\sqrt[3]{4a^4b^5} + \frac{a-c}{(b+c)^2}\)
\(= \sqrt[3]{4(4)^4(2)^5} + \frac{(4)-(-5)}{((2)+(-5))^2}\)
\(= \sqrt[3]{1024 \cdot 32} + \frac{9}{(-3)^2}\)
\(= \sqrt[3]{32768} + \frac{9}{9}\)
\(= 32 + 1\)
\(= 33\)
I think \(^3√2x + 9^3√2x - 2^3√2x\) = \(\left(-1\right)^{\frac{1}{3}}*7^{\frac{1}{3}}+40824\)
Nice solution, but I think that you have to consider that a cannot equal 0. (otherwise it wouldn't be cubic
)
I think there are actually 2.
You can tell from the discriminant, b^2 - 4ac
If b^2 - 4ac = 0, then there's 1.
If b^2 - 4ac > 0, then there are 2.
If b^2 - 4ac < 0, then there are 0.
=^._.^=
If f(0) = 0, c = 0
ax^2 + bx
Since f(1) = 2, a+b = 2.
Since f(2) = 9, 4a + 2b = 9.
a + b = 2
4a + 2b = 9
Can you solve for a and b?
Good
No problem!
yes it dose thx
Hope it helps!
because if it is a number you must type you can't copy and past numbers. sorry
