Equation 1.
20÷4=5
0×3=0
5-5=0
Equation 2.
44÷4=11
2×3=6
11-5=6
Equation 3.
56÷4=14
3×3=9
14-5=9
8) So remember the sum of all angles inside a triangle is equal to 180°. We first want to figure out the angle below the 110°.
A straight line is 180° so if we do 180 - 110 we get 70° therefore the angle below the 110° is 70°.
So we now know two angles in the triangle, 70° and 80°. Add these and we get 150°. Now we can determine the final angle (the one labelled (2x - 6)) which will be 30°.
So let’s make it into an equation:
2x - 6 = 30
Add 6 to both sides to isolate 2x:
2x = 36
Divide by 2 to get the value of x:
x = 18
10) Since the sum of all angles in a triangle is 180, and there are only 3 angles, we can determine A easily.
B = 10°
C = 160°
So to determine A:
180 - 160 - 10
Which gives us 10
Therefore A = 10°.
So the answers:
8) x = 18
9) A = 10°
I hope this helps you!!
Answer:
459
Step-by-step explanation:
First line:
clock showing 9 o'clock + clock showing 9 o'clock + clock showing 3 o'clock = 9 + 9 + 3 = 21
Second line:
The calculators look exactly the same.
3 * calculators = 30
1 calculator = 10
Third line:
1 bulb + 1 bulb - 1 bulb = 15
Since 1 bulb - 1 bulb = 0, we have 1 bulb + 0 = 15, or
1 bulb = 15
Last line:
Clock showing 9 o'clock = 9
Calculator = 10
3 bulbs = 3 * 15 = 45
Total = 9 + 10 * 45 = 9 + 450 = 459
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)