The bat costs $50 but on clearance, it is an additional 30% off.
50 - (50x.3) = x
50 - 15 = x
35 = x
The sale price of the bat is $35.
Make an equation for each statement with x being catcher's mitts and y being outfielder's gloves.
x + 10y = 239.50 and x + 5y = 134.50
Multiply the second equation through by -1 giving -x - 5y = -134.50
Combine the equations vertically
x + 10y = 239.50
-x - 5y = -134.50
giving
5y = 105
y = 21
Now substitute for y:
x + 10(21) = 239.50
x + 210 = 239.50
x = 29.50
Thus, Catchers mitts $29.50 and outfielder's gloves are $21
(2x+3y)⁴
1) let 2x = a and 3y = b
(a+b)⁴ = a⁴ + a³b + a²b² + ab³ + b⁴
Now let's find the coefficient of each factor using Pascal Triangle
0 | 1
1 | 1 1
2 | 1 2 1
3 | 1 3 3 1
4 | 1 4 6 4 1
0,1,2,3,4,.. represent the exponents of binomials
Since our binomial has a 4th exponents, the coefficients are respectively:
(1)a⁴ + (4)a³b + (6)a²b² + (4)ab³ + (1)b⁴
Now replace a and b by their real values in (1):
2⁴x⁴ +(4)8x³(3y) + (6)(2²x²)(3²y²) + (4)(2x)(3³y³) + (1)(3⁴)(y⁴)
16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
