What is 7 1/2 - 3 4/41

Mathematics

2 answers:

Radda [10]3 years ago

7 0

The LCD of 2 and 41 is 82 so we write 1/2 as 41/82:-

7 41/82 - 3 8/82

= 4 33/82 Answer

nignag [31]3 years ago

6 0

7x2+1 is 15/2. 3x41+1 is 124/41
The LCF is 82 so you would multiply everything by 82
1230/2-10,168/41
615-248=367 I might be wrong tho

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Anna [14]

$(x^3-8x^2+19x-9)\div(x-4)=x^2-4x+3\\\underline{-x^3+4x^2}\\=-4x^2+19x\\\underline{\ \ \ \ \ 4x^2-16x}\\.\ \ \ =3x-9\\\underline{\ \ \ \ \ -3x+12}\\.\ \ \ \ \ =\ 3$

$x^3-8x^2+19x-9=(x^2-4x+3)(x-4)+3$

6 0

3 years ago

Which is an example of a substitution mutation?

natka813 [3]

Answer:

1. Such a substitution could: change a codon to one that encodes a different amino acid and cause a small change in the protein produced. For example, sickle cell anemia is caused by a substitution in the beta-hemoglobin gene, which alters a single amino acid in the protein produced.

2. A - Mutations are sometimes helpful, sometimes harmful, and sometimes neutral

Step-by-step explanation:

7 0

3 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$