Answer: 
or 
Step-by-step explanation:
For this exercise you can convert the mixed number to an improper fraction:
1. Multiply the whole number part by the denominator of the fraction.
2. Add the product obtained and the numerator of the fraction (This will be the new numerator).
3. The denominator does not change.
Then:
You know that he had 30 minutes in time-out, he counted spots on the ceiling for 
minutes and the rest of the time he made faces at his stuffed tiger.
Then, in order to calculate the time Calvin spent making faces at his stuffed tiger, you need to subract 30 minutes and minutes:
or
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)
Standard form Ax + By = C
so
<span>(y - y1) = m(x - x1)
</span>given (5, -7) and m = -1/5
so
y + 7 = -1/5(x - 5)
-5y - 35 = x - 5
x + 5y = -30
answer
d. x + 5y = -30
Answer:
10p coins does he have 102
Step-by-step explanation:
given data
saves= 10p coins and 20p coins
total = 17 pounds
to find out
how 10p coins does he have
solution
we consider here no of 20p coin = x
so equation will be here
3 × 10 x +20 x = 1700
30 x + 20 x = 1700
x = 34
so that
10p coins = 3 time × 34 = 102
20p coins = 34
so that 10p coins does he have 102
Answer:
D. 1 and 20
Step-by-step explanation:
To solve the equation, factor the quadratic and set the factors equal to 0.
d^2 - 21d + 20 becomes (d-20)(d-1) since -1*-20 = 20 and -1+-20 = -21.
Set each equal to 0 and solve for x.
d-20 = 0 d = 20
d - 1 = 0 d = 1
