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blagie [28]
2 years ago
5

These prisms have different shapes as end faces

Mathematics
1 answer:
Shkiper50 [21]2 years ago
3 0

Triangle (3 sides) 5 9 6

Rectangle (4 sides) 6 12 8

Pentagon (5 sides) 7 15 10

Hexagon (6 sides) 8 18 12

b) 300 edges and 200 vertices

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William buys a book for $14.85. If he pays with a $20 bill, what will his change c be? Write an equation.
butalik [34]
<h3>Answer:  14.85 + c = 20</h3>

============================================================

Reason:

To compute the change, we subtract the amount William gives the cashier (the $20) and the price of the book ($14.85)

The change c is

c = 20-14.85

To rearrange this equation, we can add 14.85 to both sides to end up with 14.85 + c = 20

In other words, solving that equation in bold leads to c = 20-14.85 = 5.15 which is the change William is given.

----------------

Extra info (optional section)

If the cashier wanted to mentally figure out the change in his/her head, then notice the jump from 85 cents to 100 cents is 15 cents. That means we go from 14.85 to 15.00

Then the jump from $15 to $20 is an extra 5 dollars. Overall, the total change given back to William is 0.15+5 = 5.15 dollars.

4 0
2 years ago
Solve
grin007 [14]
Answer would be a (-1, ten-thirds) :)
8 0
2 years ago
Solve this r/6=5 2/3
myrzilka [38]
First, incorporate the 5 into the 2/3 fraction by multiplying the 5 by the 3 and then adding that quantity by the 2, which now makes the fraction:
17/3
Now set it up with the equation:
r/6 = 17/3
r/6×3 = 17/3×3
r/3 = 17
r/3×3 = 17×3
r = 51
4 0
3 years ago
Read 2 more answers
How much is 50 dollars
Black_prince [1.1K]
Answer
50 dollars
Explanation
4 0
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A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
Julli [10]

\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

5 0
2 years ago
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