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3 years ago

7

The area of a square garden is 81 feet. If you want to put a fence on only 3 sides of the garden, how many feet of fence will yo

u need to purchase?

1 answer:

sergeinik [125]3 years ago

4 0

Length of each side
$= \sqrt{81} \\ = 9$
Fence needed=9×3=27 ft

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HELP ME WITH IXL! SOMEONE, PLEASE HELP ME WITH IXL... THIS IS CONFUSING FOR ME. AND PLEASE PUT A EXPLANATION TO YOUR ANSWER... P

rodikova [14]

WHAT GRADE ARE YOU IN

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3 years ago

HElPPPPPPPPP PPZPZLXPLZPLZPZZZZZZZZ

monitta

Answer:

5. 43 / 6

6. 8.027778

2. -0.167

4. 1.5

Step-by-step explanation:

5. Conversion a mixed number to a improper fraction: 9 2/3 = 9 · 3 + 2 /3 = 29/3

Conversion a mixed number to a improper fraction: 2 1/2 = 2 · 2 + 1 /2 = 5/2

Subtract: 29/3 - 5/2 = 29 · 2 / 3 · 2 - 5 · 3 / 2 · 3 = 58 / 6 - 15 / 6 = 58 - 15 / 6 = 43 / 6

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4. 1 9/10 - 2/5 = 3 / 2 = 11 / 2 = 1.5

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2 years ago

there are 25 students in Hansels class and 17 of the students went on the class trip to the zoo. what percentage of students wen

drek231 [11]

It is given in the problem that out of 25 students in Hansels class 17 students went to the class trip to the zoo. From the above information it is easy to find the percentage of students that went for the class trip to the zoo.
Total number of students in the class = 25
Number of students that went for the class trip to the zoo = 17
Then
Percentage of students that went to the zoo = (17/25) * 100
= (17 * 4) percent
= 68 percent
So 68% of the students in Hansels class went for the class trip to the zoo.

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2 years ago

HOW MANY LBS ARE IN 96 OZ SHOW WORK

Andreyy89

Answer: your answer is 6 pounds

Step-by-step explanation:

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3 years ago

Read 2 more answers

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

2 years ago