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4 years ago

Can someone answer the questions pls

Mathematics

1 answer:

Aneli [31]4 years ago

5 0

Answer:

16.81 and 12.25

Step-by-step explanation:

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A line with a slope of -3 that passes through the point (-2, 6)

scZoUnD [109]

I believe the equation is y = -3x + 0

3 0

3 years ago

Mrs. Blow's chocolate chip recipe uses 8oz of chocolate chips. will she need more or less that 1lb of chocolate chips? explain

Vsevolod [243]

8 oz. is half a pound. so she will need less than a pound of chocolate.

5 0

3 years ago

У = 3х +5 y = 8х +3 What’s the ordered pair answer in decimal form?

ludmilkaskok [199]

Answer:

(0.4,6.2)

Step-by-step explanation:

set the equations equal, 3x+5=8x+3

solve for x, 5x=2, x=2/5 = 0.4

plug in x value to solve for y, y = 3*0.4 + 5, y=6.2

5 0

3 years ago

Solve the equation for x. 3(6x - 1) = 12

Law Incorporation [45]

Answer:

soln

18x-3=12

18x=15

so,x=5/6

5 0

3 years ago

Find the sum Sn below:

Hoochie [10]

We can write S as

$\displaystyle S = \sum_{k=0}^{n-1} (n-k)3^k$

and expand it as

$\displaystyle S = n \sum_{k=0}^{n-1} 3^k - \sum_{k=0}^{n-1} k\cdot3^k$

The first sum is geometric, nothing tricky:

$\displaystyle\sum_{k=0}^{n-1} 3^k = 1 + 3 + 3^2 + \cdots + 3^{n-1} \\\\ \implies 3\sum_{k=0}^{n-1} 3^k = 3 + 3^2 + 3^3 + \cdots + 3^n \\\\ \implies -2\sum_{k=0}^{n-1} 3^k = 1 - 3^n \\\\ \implies \sum_{k=0}^{n-1} 3^k = \frac{3^n-1}2$

For the second sum, you can use the same method employed in another question of yours (24494877) to find

$\displaystyle \sum_{k=0}^{n-1} k\cdot 3^k = \frac{(2n-3)3^n+3}4$

So this sum comes out to

$\displaystyle S = n\cdot\frac{3^n-1}2 - \frac{(2n-3)3^n+3}4 \\\\ S = \frac{2n\cdot3^n-2n - (2n-3)3^n-3}4 \\\\ \boxed{S = \frac{3^{n+1}-2n-3}4}$

7 0

3 years ago