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vichka [17]
3 years ago
6

I NEED HELP QUICKKKKK​

Mathematics
1 answer:
serg [7]3 years ago
5 0
I’m pretty sure it’s the last one
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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
2 years ago
u can look this up here on brainly for an answer source if you want but do the one that has the greatest points... its on google
Alla [95]

Answer:

First one is 41

Second one is 87

Step-by-step explanation:

Make the shaded part as a trapezoid which is (3+6)*10/2 you just have to get rid of 2*2. So you get 41 as the answer.

The unshaded part will be 16*8 then subtracts the shaded part which is 41.  Then you get the answer 87.

6 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
288 times 66 =?<br> Plz help
gtnhenbr [62]
19,008 I just used a calculator
4 0
2 years ago
Read 2 more answers
If x = 3 (y + 2) minus 1 what is the value of w in terms of x and y?
topjm [15]

Answer:

B

Step-by-step explanation:

edge2020

7 0
2 years ago
Read 2 more answers
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