This question was already answered: https://web2.0calc.com/questions/need-help-with-algebra_18
x^2 + (k - 10) + 36
square of a binomial = (x + b)^2 = x^2 + 2bx + b^2
The constant term is 36, so b is 6 or -6.
k - 10 = 12 or k - 10 = -12
k = 22 or k = -2
We can write everything in terms of x, the number of pages Gerald has reawd.
Gerald = x
Nash = x + 148
Josh = 2(x + 148)
x + (x + 148) + 2(x + 148) = 2724
x + x + 148 + 2x + 296 = 2724
4x + 444 = 2724
4x = 2280
x = 570
Gerald has read 570 pages.
subtract the probability that 0, 1, or 2 children don't recognize their mother's voice from 1
\(1 - (0.9)^{20} - (0.1)^1(0.9)^{19} - (0.1)^2(0.9)^{18} \approx \boxed{0.8634}\)
400,000 - 2,345 = 397,655
answers to a) to c) here: https://web2.0calc.com/questions/a-2-2-2-1-b-5-1-5-0-3-1-c-3-10-2-5#r1
\(D) \qquad \frac{6^2 - (-7)^2}{3^{-2} + 3^{-1}} = \frac{-13}{\frac{4}{9}}= \boxed{-\frac{117}{4}}\)
\(A) \qquad (-2)^2 - 2^1=4-2=\boxed{2}\)
\(B) \qquad 5^{-1} + 5^0 - 3 - 1 = \frac{1}{5}+1-4=\boxed{-2.8}\)
\(C) \qquad (3.10)^{-2} + 5 = \frac{1}{(3.10)^2}+5 \approx \boxed{5.104}\)
\(\frac{1234567890}{0987654321} \cdot 1234567890-0987654321+1234567890\)
\(\approx 1.249999 \cdot 1234567890 + 246913569\)
\(= \boxed{1790123417}\)
Let p = the number of pennies, d = the number of dimes, and q = the number of quarters. We can set up a system of equatoins:
p + d + q = 42 (eq. 1)
q = 2p (eq. 2)
d = 3p (eq. 3)
p + 10d + 25q = 567 (eq. 4)
Plug in the results of equations 2 and 3 into equation 1.
p + 3p + 2p = 42
6p = 42
p = 7
d = 3p = 21
q = 2p = 14