Let
rA--------> radius of the circle A
rB-------> radius of the circle B
SA------> the area of the sector for circle A
SB------> the area of the sector for circle B
we have that
rA=5/2 ft
rB=9/2 ft
rA/rB=5/9-----------> rB/rA=9/5
SA=(25/36)π ft²
we know that
if Both circle A and circle B have a central angle , the square
of the ratio of the radius of circle A to the radius of circle B is equals to
the ratio of the area of the sector for circle A to the area of the sector for
circle B
(rA/rB) ^2=SA/SB-----> SB=SA*(rB/rA) ^2----> SB=(9/5) ^2*(25/36)π--->
<span>SB----------- > (81/25)*(25/36)------ > 81/36------
> 9/4π ft²</span>
the answer is
<span>the measure of the sector for circle B is (9/4)π ft²</span>
First off, let's change the format of the percentage to decimal.. so 35% is just 35/100 or 0.35 and 40% is 40/100 or 0.4 and so on.

now, whatever "x" amount is, we know that x + 10 must add up to "y". x + 10 = y, because both quantities added will give the mixture amount.
and whatever 0.35x + 4 = 0.37y, because both concentrated amounts must give the 37% of the mixture.

solve for "x", to see how much of the 35% solution will be needed.
what about "y"? well, x + 10 = y.
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
A giant star is a star with substantially larger radius and luminosity than a main-sequence (or dwarf) star of the same surface temperature. They lie above the main sequence (luminosity class V in the Yerkes spectral classification) on the Hertzsprung–Russell diagram and correspond to luminosity classes II and III.
Answer:
24 way im pretty sure
Step-by-step explanation: