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Rus_ich [418]
3 years ago
11

Only need help with number 9 plz help me

Mathematics
1 answer:
Stells [14]3 years ago
8 0
20% of 98 acres

first convert 20% into a decimal value
20% = 0.20

then multiply 0.20 with 98
0.20*98 =?
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True or false. - 0.25 = - 1/4
Katena32 [7]

Answer:

true

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help please !! thank youu
VLD [36.1K]

Answer:

d = √91 ft = about 9.53 ft

Step-by-step explanation:

use pythagorean theorem which is a²+b²=c²

(the hypotnuse which is opposite side of the right angle is always going to be C in the pythagorean theorem)

a = 3    b = ?     c = 10

(3)² + b² = (10)²

9 + b² = 100

subtract 9 from both sides

b² = 91

take the square root of both sides

b = ±√91

so the side d = √91 ft = about 9.53 ft

3 0
2 years ago
Last year, Ken bought a $300 mountain bike. His bike has depreciated by 20% since he bought it. What is
trapecia [35]

Answer:

60

Step-by-step explanation:

3 0
3 years ago
Segment km is 22 cm long.<br><br> How long is the radius of circle N?
spin [16.1K]
Since the circumference meaning the whole line is 22 centimeters long then the radius would be half of that...sooo....22/2=11. The radius for N is 11cm...hope it helps :)
3 0
3 years ago
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
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