Answer:

Step-by-step explanation:
Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).
Let's calculate the common points:
![x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1](https://tex.z-dn.net/?f=x%5E2%3D%5Csqrt%7Bx%7D%5Cqquad%5Ctext%7Bsquare%20of%20both%20sides%7D%5C%5C%5C%5C%28x%5E2%29%5E2%3D%5Cleft%28%5Csqrt%7Bx%7D%5Cright%29%5E2%5C%5C%5C%5Cx%5E4%3Dx%5Cqquad%5Ctext%7Bsubtract%7D%5C%20x%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cx%5E4-x%3D0%5Cqquad%5Ctext%7Bdistribute%7D%5C%5C%5C%5Cx%28x%5E3-1%29%3D0%5Ciff%20x%3D0%5C%20%5Cvee%5C%20x%5E3-1%3D0%5C%5C%5C%5Cx%5E3-1%3D0%5Cqquad%5Ctext%7Badd%201%20to%20both%20sides%7D%5C%5C%5C%5Cx%5E3%3D1%5Cto%20x%3D%5Csqrt%5B3%5D1%5Cto%20x%3D1)
The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.
We have integrals:
![\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cint%5Climits_%7B0%7D%5E1%28%5Csqrt%7Bx%7D%29dx-%5Cint%5Climits_%7B0%7D%5E1%28x%5E2%29dx%3D%28%2A%29%5C%5C%5C%5C%5Cint%28%5Csqrt%7Bx%7D%29dx%3D%5Cint%5Cleft%28x%5E%5Cfrac%7B1%7D%7B2%7D%5Cright%29dx%3D%5Cdfrac%7B2%7D%7B3%7Dx%5E%5Cfrac%7B3%7D%7B2%7D%3D%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B3%7D%5C%5C%5C%5C%5Cint%28x%5E2%29dx%3D%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5C%5C%5C%5C%28%2A%29%3D%5Cleft%28%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B2%7D%5Cright%5D%5E1_0-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5Cright%5D%5E1_0%3D%5Cdfrac%7B2%281%29%5Csqrt%7B1%7D%7D%7B2%7D-%5Cdfrac%7B2%280%29%5Csqrt%7B0%7D%7D%7B2%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%281%29%5E3-%5Cdfrac%7B1%7D%7B3%7D%280%29%5E3%5Cright%29%5C%5C%5C%5C%3D%5Cdfrac%7B2%281%29%281%29%7D%7B2%7D-%5Cdfrac%7B2%280%29%280%29%7D%7B2%7D-%5Cdfrac%7B1%7D%7B3%7D%281%29%7D%2B%5Cdfrac%7B1%7D%7B3%7D%280%29%3D2-0-%5Cdfrac%7B1%7D%7B3%7D%2B0%3D1%5Cdfrac%7B1%7D%7B3%7D)
Given:
Violinists : Cellists = 7:3.
Number of violinists in the school orchestra = 21
To find:
The combined number of cellists and violinists.
Solution:
Let the number of violinists and cellists in the school orchestra are 7x and 3x respectively.
Then, combined number of cellists and violinists is

...(i)
Number of violinists in the school orchestra = 21

Divide both sides by 7.

Now, the combined number of cellists and violinists is
[Using (i)]

Therefore, the combined number of cellists and violinists is 30.
For this case we have the following equation:
y = 150 * (1.06) ^ t
For the first month we have:
y = 150 * (1.06) ^ 1
y = 159 $
For the second month we have:
y = 150 * (1.06) ^ 2
y = 168.54 $
For the third month we have:
y = 150 * (1.06) ^ 1
y = 178.65 $
Answer:
d. $ 159.00 + $ 168.54 + $ 178.65
Answer:
The answer is 18
Step-by-step explanation:
1.5(p+n) for p= and n =6
As we know the values of p=6 and n=6
Solve it according to the BODMAS rule. According to this rule if we have brackets or parenthesis then first we have to solve or simplify the brackets.
Put the values in the given expression:
1.5(6+6)
Now solve the parenthesis first:
1.5(12)
Now multiply 1.5 by
The answer is 18....