Answer:
Carl has 35 dimes and 90 quarters
Step-by-step explanation:
Let the number of quarters be q and the number of dimes be d
Total number of coins is 125;
Hence;
q + d = 125 •••••••••(i)
The total value of quarters present = q * 0.25 = 0.25q
The total value of dimes present = d * 0.1 = 0.1d
Adding both gives the total
0.25q + 0.1d =26 ••••••••(ii)
So we need to solve both equations simultaneously;
From i,
q = 125 - d
Substitute this into ii
0.25(125-d) + 0.1d = 26
31.25 -0.25d + 0.1d = 26
31.25 -26 = 0.25d -0.1d
5.25 = 0.15d
d = 5.25/0.15
d = 35
Recall; q = 125 - d = 125 -35 = 90
Answer:
-x
Step-by-step explanation:
8x - 9x
Factor out an x
x(8-9)
x (-1)
-x
It will be secx = 2
or, cosx = 1/2
or x = Π/3 , 5Π/3
The correct answer is: [B]: "40 yd² " .
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First, find the area of the triangle:
The formula of the area of a triangle, "A":
A = (1/2) * b * h ;
in which: " A = area (in units 'squared') ; in our case, " yd² " ;
" b = base length" = 6 yd.
" h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
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→ A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ;
= " 24 yd² " .
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Now, find the area, "A", of the square:
The formula for the area, "A" of a square:
A = s² ;
in which: "A = area (in "units squared") ; in our case, " yd² " ;
"s = side length (since a 'square' has all FOUR (4) equal side lengths);
A = s² = (4 yd)² = 4² * yd² = "16 yd² "
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Now, we add the areas of BOTH the triangle AND the square:
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→ " 24 yd² + 16 yd² " ;
to get: " 40 yd² " ; which is: Answer choice: [B]: " 40 yd² " .
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Answer:
a/1/$400
Step-by-step explanation:
240 divided by 12=20
20x10x2=$400
