How many gm equal mg?

How many gm equal mg?
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1000 mg equals 1 gm
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Description : Normal range of serum albumin is (A) 2.0–3.6 gm/dl (B) 2.0–3.6 mg/dl (C) 3.5–5.5 gm/dl (D) 3.5–5.5 mg/dl

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Description : The formula to calculate maximum urea clearance is U V× B , where U denotes (A) Concentration of urea in urine in gm/24hr (B) Concentration of urea in urine in mg/100 ml (C) Concentration of urea in blood in mg/100 ml (D) Volume of urine in ml/mt

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Daily excretion of nitrogen by an adult man is about (A) 15–20 mg (B) 1.5–2 gm (C) 5–10 gm (D) 15–20 gm

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Description : On nephelometry turbidity unit (NTU) is equal to the turbidity produced by (a) 1 mg SiO2 dissolved in 1 l of distilled water with the test being run according to absorption principle (b) 1 ... Formazin dissolved in 1 l o f distilled water with the test being run according to scattering principle.

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A ball weighing 250 gm is thrown vertically upwards with a velocity of 980 cm/sec. The time that the ball will take to return back to earth would be a.1 sec b.4 sec c.2 sec d.5 sec e.3 sec

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What weight of Na will contain the same number of atoms as are present in 2.4 gm of carbon? -Chemistry 9th

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Last Answer : 1 mole of Carbon has weight 12g. So 0.2 mole of Carbon will have weight 2.4 g. Now, 1 mole of Carbon contains 6.022 10^23 atoms So 0.2 moles of Carbon contains 1.2044 10^23 atoms. So we need to find the weight ... of Na. 1 mole of Na = 23 g So 0.2 mole of Na = 4.6 grams So, answer is 4.6 grams.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

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Last Answer : Solution :-

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Description : Find all the angles of the | |gm ABCD given in the Fig. 8.8. -Maths 9th

Last Answer : Solution :-

Two opposite angles of a ||gm are (60–x) degree and (3x –4) degree. Find the measure of each angles of the IIgm. -Maths 9th

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Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

Description : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

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