Make them count jelly beans.
<em>Greetings from Brasil...</em>
According to the statement of the question, we can assemble the following system of equation:
X · Y = - 2 i
X + Y = 7 ii
isolating X from i and replacing in ii:
X · Y = - 2
X = - 2/Y
X + Y = 7
(- 2/Y) + Y = 7 <em>multiplying everything by Y</em>
(- 2Y/Y) + Y·Y = 7·Y
- 2 + Y² = 7X <em> rearranging everything</em>
Y² - 7X - 2 = 0 <em>2nd degree equation</em>
Δ = b² - 4·a·c
Δ = (- 7)² - 4·1·(- 2)
Δ = 49 + 8
Δ = 57
X = (- b ± √Δ)/2a
X' = (- (- 7) ± √57)/2·1
X' = (7 + √57)/2
X' = (7 - √57)/2
So, the numbers are:
<h2>
(7 + √57)/2</h2>
and
<h2>
(7 - √57)/2</h2>
Hilda bought 15.11 lb of pork.
Step-by-step explanation:
Cost of Beef that Hilda bought = 4 17/20 = 4.85 lb
Cost of the total meat she bought = 19 24/25 = 19.96 lb.
To find the number of pounds of pork she bought, subtract.
Number of pounds of pork she bought = 19.96 - 4.85 = 15.11 lb.
The answer is 20in. You multiply 4 by 5 because 5x5 is 25 so the answer is 20
The zeroes of <span>f(x) = x^5-12x^2+32x can be found by factoring,
</span><span>f(x) = x^5-12x^2+32x=(x-8)(x-4)
By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.
So the zeroes of f(x) are S={4,8}</span>
