Answer: I believe this will be your answer for this.
<u>x = </u>
<u> so basically this is your solution of x equaling to 2 over 729.</u>
<u>(hope this helps!)</u>
We have the equation:
We know two points and we will use them to calculate the parameters a and b.
The point (0,3) will let us know a, as b^0=1.
Now, we use the point (2, 108/25) to calcualte b:
![\begin{gathered} y=3\cdot b^x \\ \frac{108}{25}=3\cdot b^2 \\ 3\cdot b^2=\frac{108}{25} \\ b^2=\frac{108}{25\cdot3}=\frac{108}{3}\cdot\frac{1}{25}=\frac{36}{25} \\ b=\sqrt[]{\frac{36}{25}} \\ b=\frac{\sqrt[]{36}}{\sqrt[]{25}} \\ b=\frac{6}{5} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D3%5Ccdot%20b%5Ex%20%5C%5C%20%5Cfrac%7B108%7D%7B25%7D%3D3%5Ccdot%20b%5E2%20%5C%5C%203%5Ccdot%20b%5E2%3D%5Cfrac%7B108%7D%7B25%7D%20%5C%5C%20b%5E2%3D%5Cfrac%7B108%7D%7B25%5Ccdot3%7D%3D%5Cfrac%7B108%7D%7B3%7D%5Ccdot%5Cfrac%7B1%7D%7B25%7D%3D%5Cfrac%7B36%7D%7B25%7D%20%5C%5C%20b%3D%5Csqrt%5B%5D%7B%5Cfrac%7B36%7D%7B25%7D%7D%20%5C%5C%20b%3D%5Cfrac%7B%5Csqrt%5B%5D%7B36%7D%7D%7B%5Csqrt%5B%5D%7B25%7D%7D%20%5C%5C%20b%3D%5Cfrac%7B6%7D%7B5%7D%20%5Cend%7Bgathered%7D)
Then, we can write the equation as:
Got a calculator? all you do for #1 and #2 is multiply the two numbers and that is the value for x.
These problems are cross multiplication.
For the next two:
.25x = 6
divide both sides by .25
x = 24
NEXT ONE:
.49x = 12
divide both sides by .49
x = 24.48
Answer:
305 degree
Step-by-step explanation:
As AD and CE are the diameters of circle P, so that they intersect each other at point P.
=> So that ∡CPA and ∡DPE are two vertical angles.
=> ∡CPA = ∡DPE = 93 degree
As we can see, ∡CPA = ∡CPB + ∡BPA =38 + ∡BPA
=> ∡BPA = ∡CPA - 38 = 93 - 38 = 55 degree
As P is the centre of the circle, so that ∡BPA is equal to the measure of the arc AB, equal to 55 degree.
In the circle P, the total measure of arc AEB and arc AB are 360 degree
=> arc AEB + arc AB = 360
=> arc AEB = 360 - arc AB = 360 - 55 = 305
So that the measure of arc AEB is 305 degree
Answer:
5y+5
Step-by-step explanation:
I hope this is right
