(3x - 44) / (2x + 44) = 5 / 7
x = 48
3x +2x = 240 (The total number of game cards both of them had)
[LPQ] = √28(3√28)
10 , 100 , 101 , 102 , 103 , 104 , 105 , 106 , 107 , 108 , 109 , 110 , 120 , 130 , 140 , 150 , 160 , 170 , 180 , 190 , 201 , 210 , 301 , 310 , 401 , 410 , 501 , 510 , 601 , 610 , 701 , 710 , 801 , 810 , 901 , 910 , 1000 , >>Total = 37 such numbers
29 950 - 12 ( 55 + 60) - 12 000 = taxable income
Hello Guest,
the number is beetween 8 and 6 and that's the number 7:
\(4-x=7\)
\(x=-3\)
\(4+3=7\)
So x is finally -3.
Straight
12? OR 3?
Red pens => 58+x Blue pens => 48+x Green pens => x
He sold 3/4 blue pens and 1/2 green pens.
(58+x) +1/4(48+x) + x/2 = 315
x = 140
Consider the polynomials f(x)=4x^3+3x^2+2x+1 and g(x)=3-4x+5x^2-8x^3. Find c such that the polynomial f(x)+cg(x) has degree 2.
Hello Guest!
c = 1/2
!
For what value of b is f(1)=1?
\(f(1)=3x^4-7x^3-2x^2-({\color{blue}-6})\cdot x+1=1\)
For the value of b = - 6 is f(1) = 1.
How many prizes did he buy altogether?
\(x=54+\frac{54}{6}+\frac{2\cdot 54}{6}=54+9+18=\color{blue}81\)
P W F
81 prizes did he buy altogether.
TES are the only shared letters
So there are 3+6 +6 = 15 shared permutations
Answered on the original .
Morgan starts with 4k bookmarks and Tracey starts with 3k bookmarks
Morgen give half of hers to Tracey so now Morgan has 2k bookmarks and Tracey has 5k bookmarks
BUT morgen has 9 more than Tracey which means that Mogen must have 5k-9 bookmarks
So Tracy has 2k bookmarks which is also 5k-9 bookmarks
which means that
2k=5k-9
Once you find k substitute it back to fin how many bookmarks Morgan started with.
4 missing values are: 15, 5, 11, and 8
Is is also answered herehttps://web2.0calc.com/questions/ratios_31
CX = AX / 4
BX / CX = AX / BX
BX = 6
Angle BAC => x
10x - 18 + 2x = 180
x = 16.5
Angle DAC = 2x
Find the perimeter of the triangle whose vertices are A(−1,−6), C(14,2), and B(8,−6)). Write the exact answer.
\(c=x_B-x_A=8-(-1)=9\)
\(a=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}=\sqrt{(14-8)^2+(2-(-6))^2}=10\)
\(b=\sqrt{(y_C-y_A)^2+(x_C-x_A)^2}=\sqrt{(2-(-6))^2+(14-(-1))^2}=17\)
\(P_△=a+b+c=10+17+9\)
\(P_△=36\)
What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + t^2 - t + 21.
\(f(x)=t^2 -9t - 36 + t^2 - t + 21\\ \frac{df(x)}{dx}=2t-9+2t-1=0\\ 4t=10\\ \color{blue}t=2.5 \)
The value of t = 2.5 produces the smallest value of the quadratic t^2 -9t - 36 + t^2 - t + 21.
For how many real values of x is sqrt(63 - 3*sqrt(x)) an integer?
\(f(x)=\sqrt{63 - 3\cdot \sqrt{x}}\in \mathbb Z\)
\(x\in \{81,324\} \)
\(f(x)\in \{\ 6\ ,\ 3\ \}\)
For two real values of x is sqrt(63 - 3*sqrt(x)) an integer.
Just put in x = -2 and compute ( = 2)
then put in x = -1 and compute ( = ?)
then put in x = 0 and compute (= 1) then add them together
