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In left side heart failure, the heart failure cells are seen in

In left side heart failure, the heart failure cells are seen in Correct Answer Lungs

In the left side of heart failure, the heart failure cells are seen in Lungs.
In left heart failure, the left ventricle cannot keep pace with the incoming blood from the pulmonary veins.
Heart failure cells are siderophages generated in the alveoli of the lungs of people with left heart failure or chronic pulmonary edema when high pulmonary blood pressure causes red blood cells to pass through the vascular wall.

Related Questions

Which of the following statement(s) is/are CORRECT for antigen activated effector T cells?
P. CD4+ cells make contact with macrophages and stimulate their microbicidal activity.
Q. CD4+ cells make contact with B cells and stimulate them to differentiate into plasma cells.
R. CD8+ cells make contact with B cells and stimulate them to differentiate into plasma cells.
S. CD8+ cells make contact with virus infected cells and kill them.
If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
In the question two statements are given, followed by two conclusions, I and II. Read all the conclusions and then decide which of the given conclusions logically follow/s from the given statements. Statements: All white blood cells are red blood cells. Some blood cells are red blood cells. Conclusions: I. All blood cells are red blood cells, is a possibility. II. Some red blood cells are not blood cells.
S1:The heart is pump of life.
P :They have even succeeded in heart transplants.
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S6:All this was made possible by the invention of heart-lung machine.

The Proper sequence should be:
The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :
One side of a triangle and a rectangle is the same. The third vertex of the triangle lies on the midpoint of the rectangle side opposite the common side. The rectangle sides are in the ratio 2 : 3 and the common side is the longer side of the rectangle.  What is the sin value of the angle between the common side and the equal side of the triangle?
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