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3 years ago

11

Mrs.tokalaka uses 9/10 of a can of coffe beans,each week.how many cans of beans does she use in 6 weeks? write your answer two d

ifferent ways

1 answer:

Snowcat [4.5K]3 years ago

3 0

9 x 6 = 54

so you can write it as
54/60
or = 22/30
= 11/15

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Your customer is located 100 miles from your office.

OlgaM077 [116]

Answer:

8:02 AM

Step-by-step explanation:

Find out how many minutes it takes to travel one mile:

60 divided by 45= 1.333334 minutes

It takes 1.333334 minutes to travel 1 mile, how long will it take to travel 100 miles:

1.333 times 100 = 133.3334

133.3334 rounds to 133 minutes

Add 15 minutes because you want to arrive early

133+15=148

Divide it by 60

148 divided by 60 = 2.466667

or do

148-120 (2 hours) = 28 minutes

It will take 2 hours and 28 minutes to get there

10:30-2:28= 8:02

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2 years ago

15 acres decreased by 80%

lawyer [7]

First, you have to find what 80% of 15 is. In order to do that you have to change 80% percent to decimal, which would be .80. Then, multiply .80 and 15 to get 12. Subtract 12 from 15 and your answer is 3.

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2 years ago

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, $c=12$ .

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1 year ago

Need help with this question. Thank you!

Paul [167]

Answer:

61.2

Step-by-step explanation:

Distribute the figure into three shapes. Triangle, and two rectangles.

Find the area.

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3 years ago

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Which expressions are equivalent to

andrew-mc [135]

Answer

30

Step-by-step explanation:

the answer is :

90 - 60 = 30

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2 years ago

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