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3 years ago

Help please ! Please

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

15, 8, 25 The teacher payed 0.40 per book.

Step-by-step explanation:

4/10 = 2/5 Simplify the 4/10 to find the unit rate.

15: 2/5 = 6/x Set up your proportion.

2x =30 Cross multiply

/2 /2 Divide both sides by 2 to isolate x

x =15

8: 2/5 = x/20 Set up proportion

40 =5x Cross multiply

/5 /5 Divide both sides by 5 to isolate x

8 =x

25: 2/5 = 10/x Set up proportion

2x =50 Cross multiply

/2 /2 Divide both sides by 2 to isolate x

x =25

Graph: y=mx+b To graph, you will need to use y=mx+b format

y= 2/5 x+2 2/5 is your unit rate. Your rate changes every 2 books bought. To graph, start at (0,2) and go up 2 and to the right 5, plot. Continue moving up 2 and to the right 5 to graph.

The teacher payed 0.40 per book because 2/5 simplifies to 0.40.

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I need help please............

Anettt [7]

A) x=4
B) x= -3
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D) x=24

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3 years ago

Find the volume of the solid. Round to the nearest tenth, if necessary.

puteri [66]

Answer:

I think it is correct

Step-by-step explanation:

length=10m

height=12

breadth = 3m

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3 years ago

1/3x+5=2 using fraction busters

svet-max [94.6K]

The answer is 9.
1/3 x + 5=2
The fraction buster is 3/1
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5/1 times 3/1=15/1
2/1 times3/1= 6/1
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3 years ago

Evaluate the infinite sum:

satela [25.4K]

Consider the k-th partial sum,

$S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}$

More compactly,

$\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}$

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as k goes to infinity. We have

$\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}$

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |x| < 1, we have

$\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n$

Differentiating both sides gives

$\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

also valid for |x| < 1. Take x = 1/π and you get the sum you want to compute.

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3 years ago

On Saturday you spent $33, give me $15 to a friend, and receive $20 for mowing your neighbors lawn. You have $21 left. Use two m

Lemur [1.5K]

Answer:

Step-by-step explanation:

M = 33+15+21-20

M = 49

Day started with $49

#1. 49-33=16

#2. 16-15=1

#3. 1+20=21

So basically if you do a step by step process you will most likely get $49

6 0

3 years ago