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kondaur [170]
3 years ago
14

Robert bought $1.60 worth of stamps with 20 coins all in nickels and dimes. how many nickels and dimes did he use?

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
3 0
Nickels are 5 cents, and dimes are 10 cents.
(By luck..) eventually you can have 12 dimes, meaning you have 120 cents, or $1.20 ... 20-12=8.. meaning you used 12 dimes and are missing 8 more coins. which means you can have 8 nickels which is 40 cents. 120+40=160 cents or $1.60
Hope this helped.

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Can anyone tell me what can i do with these mice baby ??​
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Need help with this question! Will attach pic! A satellite is to be put into an elliptical orbit around a moon as shown below.Th
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An equation for the ellipse ( standard form ):
x²/a²  +  y²/b² = 1
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4 0
3 years ago
Which salt produces the least salt hydrolysis in an aqueous solution?
Elodia [21]

Answer:

Potassium Iodide

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I hope this helps. If you have any more questions, please feel free to post them and someone will be able to help you, whether it's myself or others. Please leave a like, rating, and if possible, Brainliest. Have a great day!

6 0
3 years ago
The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo
PolarNik [594]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY =  \sqrt{x^2+y^2}

Therefore, the length between points AB is

length AB = \sqrt{(-4 - (-2))^2+(-1-3)^2}

= \sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2}  = \sqrt{20}  = 4.472 units

Similarly, length BC is given by

Length BC =  \sqrt{(-4)^2+1^2} =  \sqrt{17} = 4.123 units

Length CD = \sqrt{(-2)^2+5^2} =  \sqrt{29} = 5.385 units

Length DA = \sqrt{(8)^2+(-2)^2} =  \sqrt{68} = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A =  \sqrt{(7)^2+(0)^2} =  \sqrt{49} =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

 S_{\bigtriangleup} = (\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|

= (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

S_{{\bigtriangleup}ABC} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|

= =(\frac{1}{2})|-4-6 -8|= 9 units^2

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

S_{{\bigtriangleup}ACD'} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2} = 10.5 units²

Area of polygon =   S_{{\bigtriangleup}ABCD'} = S_{{\bigtriangleup}ABC} + S_{{\bigtriangleup}ACD'} = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

8 0
4 years ago
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