Answer:
independent variable: the different amount of water
Explanation:
The independent variable would be the <u>different amount of water supplied to the pea plant clones.</u>
<em>The independent variable is the variable inputted by the researcher during an experiment whose value directly affects the dependent variable. It is the variable that is manipulated during experiments in order to see the kind of effects they produce on another variable - the dependent variable.</em>
In this case, <u>the amount of water supplied to each pea plant would directly affect the height of the pea plants.</u> Hence, the amount of water supplied to the pea plants is the independent variable while the height of the plants is the dependent variable.
Answer:
A and C
Explanation:
A. because when we humans are burning maybe our refuse or garbage it will go up to the ozone layer causing air and then it will then result to carbon dioxide.
The answer is Monosaccharide
Answer:
D. Molecules called pumps function to control active transport.
Explanation:
Active transport is one of the two types of transport (the other being passive transport). It is the type of transport in which substances move against their concentration gradient i.e. from a lower concentration to a higher concentration. Hence, due to this, energy input in form of ATP is required by the cell.
However, the process of active transport makes use of certain molecules called PUMPS to control it. The pumps moves ions/molecules against their concentration gradient. An example is the sodium-pottasium pump which moves sodium ions from a region of low conc. to a region of high conc.
I think the priority action is to monitor Bowel movements.
Kayexalate causes potassium to be exchanged for sodium in the intestines and excreted through bowel movements. If the client does not have stools, the drug cannot work properly. Blood pressure and the urine output are not of primary importance. The nurse would already expect changes in T waves with hyperkalemia. Normal serum potassium is 3.5 to 5.5 mEq/L.
