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Biology

General

Easy

Question

Both sickle cell anemia and Hunington’s chorea are :

pollutant-induces disorders
virus-related diseases
bacteria-related diseases
congenital disorders

The correct answer is: congenital disorders

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HIV that causes AIDS, first starts destroying

HIV that causes AIDS, first starts destroying

A very much publicised treatment method ‘DOTS’ is being adopted for the cure of

A very much publicised treatment method ‘DOTS’ is being adopted for the cure of

Sickle cell anaemia is due to

Sickle cell anaemia is due to

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In figure, $stack A B with rightwards arrow on top equals 3 stack i with minus on top minus stack j with minus on top comma stack A C with rightwards arrow on top equals 2 stack i with minus on top plus 3 stack j with minus on top$ and $stack D E with rightwards arrow on top equals 4 stack i with minus on top minus 2 stack j with minus on top$ then the area of the shaded region in square units is

In figure, $stack A B with rightwards arrow on top equals 3 stack i with minus on top minus stack j with minus on top comma stack A C with rightwards arrow on top equals 2 stack i with minus on top plus 3 stack j with minus on top$ and $stack D E with rightwards arrow on top equals 4 stack i with minus on top minus 2 stack j with minus on top$ then the area of the shaded region in square units is

In the figure, $stack D E with bar on top divided by divided by stack B C with bar on top$ and ar $left parenthesis triangle A D E right parenthesis equals a r invisible function application left parenthesis capital delta B E D right parenthesis$ The value of $fraction numerator B D over denominator A B end fraction equals$ ?

In the figure, $stack D E with bar on top divided by divided by stack B C with bar on top$ and ar $left parenthesis triangle A D E right parenthesis equals a r invisible function application left parenthesis capital delta B E D right parenthesis$ The value of $fraction numerator B D over denominator A B end fraction equals$ ?

The median AD of the triangle ABC is bisected at E and BE meets AC at F. If AF:AC =1/k then the value k

The median AD of the triangle ABC is bisected at E and BE meets AC at F. If AF:AC =1/k then the value k

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In the figure, $blank stack A B with bar on top divided by stack C D with bar on top divided by stack E F with bar on top$ If AB = 6 cm, BE = 4cm, CD = 10cm, then EF + EC = ... Cm.

In the figure, $blank stack A B with bar on top divided by stack C D with bar on top divided by stack E F with bar on top$ If AB = 6 cm, BE = 4cm, CD = 10cm, then EF + EC = ... Cm.

In the figure, $angle equals angle A C E D$ , AB = 9cm, AD = 7cm, CD = 8cm and CE = 10cm Then DE = ?

In the figure, $angle equals angle A C E D$ , AB = 9cm, AD = 7cm, CD = 8cm and CE = 10cm Then DE = ?

In the figure, $stack A B with bar on top divided by divided by stack Q R with bar on top blank$ If AB = 3cm, PB = 2cm and PR = 6cm then QR = ? cm.

In the figure, $stack A B with bar on top divided by divided by stack Q R with bar on top blank$ If AB = 3cm, PB = 2cm and PR = 6cm then QR = ? cm.

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In the figure, $stack A B with bar on top divided by divided by stack C D with bar on top$ and $stack A C with rightwards arrow on top intersection stack B D with bar on top equals 0 blank$ If OA = 3x – 1, OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC =....units.

Hence AC=12

In the figure, $stack A B with bar on top divided by divided by stack C D with bar on top$ and $stack A C with rightwards arrow on top intersection stack B D with bar on top equals 0 blank$ If OA = 3x – 1, OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC =....units.

Hence AC=12

Two line segments $stack A B with bar on top$ and $stack C D with bar on top$ intersect at E such that $triangle A C E tilde operator triangle B D E times$ If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

Hence x=2

Two line segments $stack A B with bar on top$ and $stack C D with bar on top$ intersect at E such that $triangle A C E tilde operator triangle B D E times$ If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

Hence x=2

In the figure, AB = BC = CD = DE = EF and AF = AE Then $angle A$ = ?

In the figure, AB = BC = CD = DE = EF and AF = AE Then $angle A$ = ?

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In the figure, AB = AC, $angle A equals 48 to the power of ring operator end exponent$ and $angle A C D equals 18 to the power of ring operator end exponent$ Then

Hence BC=CD

In the figure, AB = AC, $angle A equals 48 to the power of ring operator end exponent$ and $angle A C D equals 18 to the power of ring operator end exponent$ Then

Hence BC=CD

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of DABC Then x° + y° =

Hence x+y>180^o

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of DABC Then x° + y° =

Hence x+y>180^o

In $triangle A B C comma text if end text angle A plus angle B equals 84 to the power of ring operator end exponent comma angle B plus angle C equals 146 to the power of ring operator end exponent$ then $angle B equals$

In $triangle A B C comma text if end text angle A plus angle B equals 84 to the power of ring operator end exponent comma angle B plus angle C equals 146 to the power of ring operator end exponent$ then $angle B equals$

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