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4 years ago

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Lora's gymnastics class practices floor exercises every other day.The class practices on the balance beam every third day,and th

e uneven bars every fourth day.Today is march 10,and the class practiced all three events.How many more times,before June 1,will the class practice all three on the same day.

1 answer:

sladkih [1.3K]4 years ago

3 0

The 12th decades every other day every three days and four days all have 12 in common I think

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Please Help Me! I need to pass.

Andru [333]

42 is the answer

Explanation:
50% of 80 is half of 80 which is 40

Then you would add 40 with 5% of 40

What you would do then is 5% times 40 then you would change the 5% into a decimal which is 0.05 and multiply it by 40 , which would get u 2

Then you would add the price with tax
40+2 = 42

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3 years ago

The quarterback completed 11 passes out of 15 pass

levacccp [35]

Answer:

175-176 passes are completed in a season

Step-by-step explanation:

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3 years ago

Please Help Me With This.

Romashka-Z-Leto [24]

Answer:

A is $0.45 B is 2 ounces

Step-by-step explanation:

A when you divide the cost by the ounces or $5.40 by 12 ounces you get $0.45 per ounce.

B $0.45 times 2 equals 90 cents and you cannot go over 1 dollar.

C you should be to do along with D with this information.

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3 years ago

Which step is the same when constructing an inscribed square and an inscribed regular hexagon?

Colt1911 [192]

After the construction of a circle, we have to "Set the compass to the radius of the circle" so, option C is correct.

<h3>What is a regular hexagon?</h3>

A regular hexagon is defined as a closed shape consisting of six equal sides and six equal angles. The sum of the measure of angles of a regular hexagon is 120 degrees.

Steps to create an inscribed hexagon:

1: The structure needs to adjust the box thickness towards that radius.

2: Afterward moves around the outside of the circular path to just produce the 6 vertices of that similar hexagon.

"Set the compass to the radius of the circle" so, option C is correct.

Thus the above answer is correct.

Learn more about inscribed hexagons here:

brainly.com/question/21502832

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6 0

2 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago