North component: y = 15.0 m * sin 55.0º = 12.3 m West component: x = 15.0 m * cos 55.0º + 7.00 m = 15.6 m
so his heading, measured from West, is Θ = arctan(y/x) = arctan0.787 = 38.2º N of West
and measured from North is φ = arctan(x/y) = arctan(1.27) = 51.8º W of North
Hope this helps!
I'm not sure what YOU already know about photosynthesis and cellular respiration.... however, you SHOULD know that the two are beautifully linked to one another!
Photosynthesis equation:
6CO2+6H2O -> C6H12O6+H2O
Cellular respiration
C6H12O6+H2O -> 6CO2+6H2O
Notice something? The substrates of one equation are the products of the other! We rely on plants for their photosynthetic reactions - and plants benefit from us (not really because there is already a TON of CO2 in the atmosphere) from our cellular respiration
***we like their oxygen and they like our carbon dioxide!
:) I hope that helps! Let me know if you need any more elaboration!
The true answer of your question is :
OH : HYDROXYL GROUP
NH2 : AMINO GROUP
CH3 : METHYL GROUP ( but in rather broader terms, that functional group of formula CnH2n+1 where n is an integer is called ALKYL GROUP where by substituting n by 1,2,3... we obtain methyl for n = 1, ethyl for n = 2, and propyl for n = 3 )
COOH : CARBOXYL GROUP is the correct answer since carbonyl is characterized by the presence of functional group C=O in general the formula of the compound would be
R-C=O-R’ where R and R’ are alkyl groups like methyl for example. However the carboxyl group could be thought of as a summation of carbonyl + hydroxyl ( CO + OH ) resulting thus in COOH.
I hope you’ll understand everything, anyway if not i’m always here to help. ♥️
Magnesium Phosphate is the answer
Answer No 1:
The set up of the punnet square is shown in the attached diagram. As the alleles assort independently hence the gametes formed will be OT,Ot,OT,Ot and OT,Ot,oT,ot. These will be the outcomes of the possible gametes formed. When these gametes are cross bred, the results are shown in the diagram attached.
Answer No 2:
The outcomes of each possible genotype are:
OOTT = 2/16
OOTt = 4/16
OOtt = 2/16
Oott = 2/16
OoTT = 2/16
OoTt = 4/16
Answer No 3:
The likelihood of each possible offspring phenotype is:
Orange petals with tall stem and orange petals with small stems present in ratio 12:4 i.e. 3:1.
