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FinnZ [79.3K]
2 years ago
7

#5 Antonio and Abby had the same nimber of paperclips. After Antonio gave 30 paperclips to Abby, Abby had twice as many papercli

ps as Antonio. How many paperclips did they have in all?
#6 Emily and Jasmine had the same number of stamps. After Emily gave 42 stamps to Jasmine, Jasmine had twice as many stamps as Emily. How many stamps did Jasmine have in the end?

#7 Elena had 60 colored pencils. Lucy had 26 colored pencils. How many pencils must Elena give to Lucy so that Elena will have 4 more colored pencils than Lucy?

Show how you answered please
Mathematics
1 answer:
Kay [80]2 years ago
8 0

Answer:

#5: 180 paperclips in total.

#6: 126 stamps in total.

#7: Elena should give Lucy 15 colored pencils.

Step-by-step explanation:

This explanation solves each question by setting a single unknown, x.

<h3>#5</h3>

Let x the initial number of paperclips of Antonio. That should also be the number of Abby's paperclips.

Initially:

  • Antonio: x paperclips;
  • Abby: x paperclips.

Antonio gives 30 paperclips to Abby. After that,

  • Antonio: (x - 30) paperclips;
  • Abby: x + 30 paperclips.

Abby now possess twice as many paperclips as Antonio does. In other words,

2(x - 30) = x + 30.

By the distributive property:

2x - 60 = x + 30.

Substract x - 60 from both sides

x = 30 - (-60) = 90.

Both Antonio and Abby initially possess 90 paperclips. That's 180 in total.

<h3>#6</h3>

Similarly, let x be the number of Emily's stamps. That should be the same as the number of Jasmine's stamps.

Initially:

  • Emily: x stamps;
  • Jasmine: x stamps.

After Emily gives 42 stamps to Jasmine:

  • Emily: x-42 stamps;
  • Jasmine: x+42 stamps.

Jasmine now possesses twice as many stamps as Emily does. In other words,

2(x-42) = x+42.

x = 42 + 2\times 42 = 126.

Jasmine used to possess 126 stamps. Now she possesses 126 + 42 = 168 stamps after receiving 42 stamps from Emily.

<h3>#7</h3>

Let the number of pencils that Elena needs to give to Lucy be x.

Initially:

  • Elena: 60 pencils;
  • Lucy: 26 pencils.

After Elena gives x pencils to Lucy:

  • Elena: 60 - x pencils;
  • Lucy: 26 + x pencils.

Elena should now possess four more pencils than Lucy does. In other words,

\underbrace{60 - x}_{\text{Elena's}} = \underbrace{(26 + x)}_{\text{Lucy's}} +4.

2x = 30.

x = 15.

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3 years ago
1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than
zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)

C(x, y) = (17, –1.5)

8 0
3 years ago
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