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- What is the medical term for a plexus of capillaries in the . . . - Socratic
The medical term for a plexus of capillaries in the nephron is glomerulus The glomerulus is a network of capillaries located at the beginning of a nephron in the kidney It serves as the first stage in the filtering process of the blood carried out by the nephron in its formation of urine A glomerulus along with its surrounding Bowman's capsule constitute a renal corpuscle ( basic
- Question #7e2d2 - Socratic
During filtration, some of the particles pass through the filter while others, being larger, are trapped But no molecules are changed in structure in any way, which is the primary requirement of a chemical change
- Questions asked by - Socratic
What is the difference between gravity filtration and vacuum filtration? When should vacuum filtration be used? What is the difference between copper (II) sulphate trihydrate, copper (II) sulphate pentahydrate, and anhydrous copper (II) sulphate? Methane, CH4 , the major component of natural gas burns in air to form CO2 and H2O
- Site Map - Separating Mixtures Questions and Videos | Socratic
When should vacuum filtration be used? What is the difference between copper (II) sulphate trihydrate, copper (II) sulphate pentahydrate, and anhydrous copper (II) sulphate?
- How do you find the circumference of the sun in feet? - Socratic
The circumference of the equatorial great circle of the Sun is 1 43545 E+10 = 14354 5 million feet I understand circumference of the Sun as the circumference of the solar disc that is nearly the circumference of the Sun's equator Equatorial radius of the Sun is 696342 km, nearly The circumference of the equator = 2pi X 696342 = 4375246 km =4375246 1 609344=2718652 miles =2718652 X 5280 feet
- Question #cbd67 - Socratic
Explanation: Note that the denominator is: #x^2+2x+1 = (x+1)^2# so we can substitute #t=x+1#, #x=t-1#, #dx=dt# and have: #int (x^3dx) (x^2+2x+1) = int ( (t-1)^3dt) t
- Question #21b50 - Socratic
1 3* (x^2+x)^ (3 2)-1 8* (2x+1)*sqrt (x^2+x)+1 16ln (2x+1+sqrt (4x^2+4x))+C int xsqrt (x^2+x)*dx =int xsqrt ( (x+1 2)^2-1 4)*dx =int (x+1 2-1 2)sqrt ( (x+1 2)^2-1 4
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