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Svetllana [295]
3 years ago
11

At a recycling center, 1 out of every 5 bottles collected is a glass bottle. If the recycling center collects 400 bottles, how m

any bottles are glass bottles?
Mathematics
1 answer:
Kryger [21]3 years ago
6 0

。☆✼★ ━━━━━━━━━━━━━━  ☾  

400/5 = 80

There are 80 glass bottles

Have A Nice Day ❤    

Stay Brainly! ヅ    

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Three brothers have ages that are consecutive even integers the product of the first & third boys ages is 20 more than twice
Agata [3.3K]
Let's say X is the first brother's age
x+2 is second brother, and x+4 is the third brother
Since the product of the first and third boy x(x+4) is 20 more than twice (20+x(x+2), our equation is x(x+4)=20+2(x+2)
Simplify from there and see what you get.
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3 years ago
my son had his 18 months visit to the doctor last week.the nurse took several measurement while we were there.his length(height)
nikitadnepr [17]

18 months ……33inches, weight 25.5 pounds

When he was born = 9 months…20 inches, weight 8, 4 pounds

Every 9 months his height growth is 33-20 = 13 inches, his weight is 25.5-8.4=17.1 pounds 

<span>Pounds.  This means, for every 9 months, his growth is 13 inches and his weight is 17.1 pounds. For example we can determine his growth in 3 years.</span>

<span>3 years = 36 months, 36= 9*4, his height will be 4*13 inches= 52 inches, his weight will be 4*17.1 = 68.4 pounds </span>

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2 years ago
Solving the whole thing
erik [133]
Let
x---------------> distance from people living to the city center

we Know that
Zone 1 covers people living within three miles of the city center
Zone 1 ------------> [x < 3 miles]

Zone 2 covers those between three and seven miles from the center
Zone 2 ------------> [ 3 <= x < = 7  miles]

Zone 3 covers those over seven miles from the center
Zone 3 ------------> [  x > 7  miles]

<span>calculate the distance between two points to find the value of x
</span>
case A) point (0,0)  point (3,4)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(4-0)² +(3-0)²]------> √[16+9]
x=√25-------------> x=5 miles

the answer Part A)
people living in (3,4)
x=5 miles -------------> covers Zone 2 [ 3 < =x <= 7  miles]


case B) point (0,0)  point (6,5)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(5-0)² +(6-0)²]------> √[25+36]
x=√61-------------> x=7.81 miles

the answer Part B)
people living in (6,5) 
x=7.81 miles -------------> covers Zone 3 [  x > 7  miles]

case C) point (0,0)  point (1,2)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(2-0)² +(1-0)²]------> √[4+1]
x=√5-------------> x=2.23 miles

the answer Part C)
people living in (1,2) 
x=2.23 miles -------------> covers Zone 1 [ x < 3  miles]

case D) point (0,0)  point (0,3)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(3-0)² +(0-0)²]------> √[9]
x=√9-------------> x=3 miles

the answer Part D)
people living in (0,3) 
x=3 miles -------------> covers Zone 2 [ 3 < =x <= 7  miles]

case E) point (0,0)  point (1,6)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(6-0)² +(1-0)²]------> √[36+1]
x=√37-------------> x=6.08 miles

the answer Part E)
people living in (1,6) 
x=6.08 miles -------------> covers Zone 2 [ 3 < = x <= 7  miles]

4 0
3 years ago
Anyone knows how to do questions 7 and 8? 15 pts!!
Oksi-84 [34.3K]
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
3 0
3 years ago
Read 2 more answers
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