Answer:
your answer is 1,-2
Step-by-step explanation:
Answer:
In general, IBM MQ object names can be up to 48 characters long
Answer:
r= 2 √(15/π)
Step-by-step explanation:
Given that :
Area of circle =60sq cm
radius=√60/π
![Area=\pi r^2\\60=\pi r^2\\\\\sqrt{\frac{60}{\pi}}=r\\\\\sqrt{\frac{15\times 4}{\pi}}=r\\2\sqrt{\frac{15}{\pi}}=r](https://tex.z-dn.net/?f=Area%3D%5Cpi%20r%5E2%5C%5C60%3D%5Cpi%20r%5E2%5C%5C%5C%5C%5Csqrt%7B%5Cfrac%7B60%7D%7B%5Cpi%7D%7D%3Dr%5C%5C%5C%5C%5Csqrt%7B%5Cfrac%7B15%5Ctimes%204%7D%7B%5Cpi%7D%7D%3Dr%5C%5C2%5Csqrt%7B%5Cfrac%7B15%7D%7B%5Cpi%7D%7D%3Dr)
Hence the radius can be written as 2 √(15/π)
Answer:
Do you have the figure Friend?
If yes please attach it too!
Area of part 1:A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2Area of part 2:A2=length*width=(8+2)*1=10 ft2Area of part 3:A3=length*width=8*2=16 ft2
Total shaded area : A1+A2+A3=41 ft2Total area of the reactangle : A=16*8=128 ft2Total area of the nonshaded region : 128-41=87 ft2