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3 years ago

Please help me with this question

Mathematics

2 answers:

sammy [17]3 years ago

8 0

7 1/8 is you answer.

Hope this helped!!

erastovalidia [21]3 years ago

7 0

Here, 3/8 * 3/8 = 9/64

In short, Your Answer would be: 9/64

Hope this helps!

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(3x – 1) : (x - 2) = x – 2x (x—4)<br> срочно, сор.

RSB [31]

$\frac{3x-1}{x-2}=x-2x(x-4) \frac{3x-1}{x-2}=x-2x^{2} +8x \\\frac{3x-1}{x-2}=-2x^{2} +9x3x-1= (-2x^{2} +9x)(x-2)\\3x-1= -2x^3+4x^{2} +9x^{2} -18x\\3x-1=-2x^3+13x^{2} -18x\\2x^3 -13x^{2} +18x+3x -1=0\\2x^3 -13x^{2}+21x-1=0\\ x =0.05 /or/x=3.7$

ok done. Thank to me :>

6 0

2 years ago

1) write down a different ratio equivalent to 12 : 3

Alina [70]

1) 24:6
2) 60:48
I just multiplied by 2

7 0

3 years ago

Read 2 more answers

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours

ella [17]

Answer:

137

Step-by-step explanation:

Use the model A = Pe^(kt), where P is the initial number and k is the growth constant. Then, in case 1,

300 = Pe^(k*3)

and in case 2,

4000 = Pe^(k*10). We need to find P and e^k.

300 = Pe^(k*3) can be rewritten as (300/P) = [e^k]^3, which in turn may be solved for e^k:

∛(300/P) = e^k

Now we go back to 4000 = Pe^(k*10) and rewrite it as (4000/P) = [e^k]^10. Substituting ∛(300/P) for e^k, we obtain:

(4000/P) = (300/P )^(10/3)

which must now be solved for P. Raising both sides by the power of 3, we get:

(4000/P)^3 = (300/P)^10. Therefore,

4000^3 300^10

------------- = -------------

P^3 P^10

which reduces to:

4000^3 300^10

------------- = -------------

1 P^7

------------ = -----------

4000^3 300^10

or:

300^10

P^7 = ---------------- = 9.226*10^13 = 92.26*10^12

4000^3

Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or

P = 137.26

The initial number of bacteria was 137.

7 0

3 years ago

A machine makes two components; call them types A and B. It takes 250 seconds to switch production between the component types.

Alexus [3.1K]

Answer:

The average inventory depends on the manufacturing batch size.

Its formula is:

$Inventory=\frac{n}{3n+250}$

Step-by-step explanation:

The question is incomplete.

Question complete:

<em>"A machine makes two components, call them type A and B. It takes 250 seconds to switch production between the component types. During that time no production occurs. When in production, each unit of A or B requires 0.5 seconds to be completed. The two components, A and B are combined in an assembly process to make a final product, call it C. The assembly step can combine the two components into 1 unit every 2 seconds. What is the average inventory of B components?"</em>

We have two steps in this process: the manufacturing of A and B and the assembly process.

We are not told about the batch size, so we will treat that as a variable called n. This represents the amount of units (of A or B) that are manufactured per cycle.

Time (seconds) to manufacture n units of A and n units of B:

$T_m=n*0.5+n*0.5+250=n+250$