Answer:
The length of the square is 6 units.
Step-by-step explanation:
Given that all sides of the square are the same and the area of square formula is A = length×breadth. So let's make x as the sides :







10 cakes can be made
Explanation:
Convert 7 1/2 to an improper fraction
15/2 divided by 3/4
Flip the second fraction and multiply instead of divide
15/2 x 4/3
multiply the tops: 15 x 4 = 60
Over the bottoms multiplied together 2x3=6
60/6=10
therefore 10 cakes can be made
Answer:
Step-by-step explanation:
We can eliminate options C and D quite quickly because factoring either one would have one equation be y = x and the other either y = 72x - 96 or y = 72x + 108. Neither of these added to x will give either option A or B
so that leaves E.
72x² – 204x + 144 = 0
we can find the zeros
x = (204 ± √(204² - 4(72)(144))) / (2(72))
x = (204 ± 12) / 144
x = 1.5
x = 4/3
so the equation has factors
(x - 1.5)(x - 4/3)
and therefore also a factor in their product
x² - (
)x + 2
which is 1/72 of the original quadratic
so all factors of E are
72(x - 1.5)(x - 4/3)
now we need to distribute factors of 72 among the other two factors so that when we add them together the x¹ terms are either 16 or 17 and the x⁰ terms sum to -12. let "a" and "b" be the factors of 72.
ab = 72
a = 72/b
-1.5a - 4b/3 = -12
1.5a + 4b/3 = 12
1.5(72/b) + 4b/3 = 12
108/b + 4b/3 = 12
324/b + 4b = 36
324 + 4b² = 36b
81 + b² = 9b
b² - 9b + 81 = 0
b = (9 ±√(9² - 4(1)(81))) / 2(1))
b = (9 ± √-243) / 2
as both of these roots are imaginary numbers
there is no valid solution to this problem as posed
IF we allow a slight edit to answer B, we can factor 72 into 9•8
y = 8(x - 1.5) y = 9(x - 4/3)
y = 8x - 12 y = 9x - 12
so the sum of the two would be 17x - 24
Answer:
True
Step-by-step explanation:
This is a part of both female/ male. The male has sperm when released, and the female produces a egg which lead to reproduction.
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8
Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0)
Winning once W1 is equal to L4, winning zero times is losing 5 times.
p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5);
p (W1) + p(W0) = p(L4) + p(L5)
p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5
p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>