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3 years ago

(8/9)^2 I need to evaluate this

Mathematics

2 answers:

GaryK [48]3 years ago

8 0

The answer is 64/81. We can simplify it anymore so that would our final answer

Hope this helps

tigry1 [53]3 years ago

7 0

(8/9)^2 is just 8/9 squared, or 8/9 * 8/9. This is 64/81, and cannot be reduced. This is therefore our answer.

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The price paid for a $250 table after a 30% discount is applied

Vladimir [108]

Answer:

GAFD

Step-by-step explanation:

250*0.3 is 75

250-75 175

so D

GAFD

Hope this helps plz hit the crown :D

8 0

2 years ago

Read 2 more answers

A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rat

Ivanshal [37]

We are told that a population of insects grows exponentially and we are given a table of data about insect population growth. We are asked to find population of insects at the end of week 11.

The initial insect population is 20 and at the end of 1st week population increases to 30.

Let us find growth percentage of insect population,

$\text{Growth percentage}=\frac{\text{Difference}}{Actual} \cdot100$

$\text{Growth percentage}=\frac{\text{30-20}}{20} \cdot100$

$\text{Growth percentage}=\frac{\text{10}}{20} \cdot100$

$\text{Growth percentage}=0.5 \cdot100=50$

We can see that insect population is growing at rate of 50% per week.

Now let us write an exponential function for our population.

$P(w)=20(1+0.50)^{w}$ , where P(w) represents population at the end of w weeks.

Let us substitute w=11 in our function to find insect population at the end of 11 weeks.

$P(11)=20(1+0.50)^{11}$

$P(11)=20(1.50)^{11}$

$P(11)=20\cdot 86.49755859375$

$P(11)=1729.951171875\approx 1730$

Therefore, population of insects at the end of 11th week will be 1730.

7 0

3 years ago

Use the surface integral in Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus

Tom [10]

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