The only whole numbers that multiply together to equal 22 are 1 x 22, and 2 x 11.
Unfortunately, neither of these pairs add OR subtract to equal 11.
Answer:
x^3-4x^2+25x-100
Step-by-step explanation:
We have 2 roots -5i and 4. Using the conjugate root theorem for the root -5i, we know that 5i should be another root
Combining the roots into a polynomial gives us:
(x+5i)(x-5i)(x-4)
Expand:
(x^2 + 25)(x-4)
x^3-4x^2+25x-100
Answer:
Scale factor for the drawing from actual park = 
Area of the actual park = 1200 cm²
Step-by-step explanation:
Length of the rectangular city park = 5 cm
Width of the rectangular park = 6 cm
Using scale factor 1 cm = 20 meters
Scale factor = 

Actual length = 5 × 20 = 100 meters
Actual width = 6 × 20 = 120 meters
Area of the rectangular park = length × width
= 100 × 120
= 1200 square meters
Therefore, Scale factor from actual length to the length in drawing = 1 : 20
Area of the rectangular park = 1200 square feet
Answer:
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C
Step-by-step explanation:
∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗
Let 1st=arctan(x)
And 2nd=1
∫▒〖arctan(x).1 dx=arctan(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗
As we know that
derivative of arctan(x)=1/(1+x^2 )
∫▒〖1 dx〗=x
So
∫▒〖arctan(x).1 dx=arctan(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1
Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now
Let 1+x^2=u
du=2xdx
Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get
1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)
1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)
1/2 ∫▒(2xdx/u) =1/2 ln(u)+C
1/2 ∫▒(2xdx/u) =1/2 ln(1+x^2 )+C
Putting values in Eq1 we get
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C (required soultion)
Ok so there are 52 cards in a deck if you take all the cards that have the number 6 which is only 4 cards so 52-4 = 48 cards without any 6 so the ratio is 48/52
chances.
If you're trying to take any card that is greater than 6 its a little more complex there is T,2,3,4,5,6,7,8,9,10,B,Q,K so you have to tack away 6,7,8,9,10,B,Q,K from 4 rows so 8 x 4 =32 cards 52-32= 20 so the probability of not having any cards greater than 6 is 20/52