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we look at the first equation, and we know that having 18 for 3 brass trinkets, or 6 for each, results in us still having steel trinkets. subtracting 18 from 48, we get thirty. dividing 30 by 10 we get 3. A brass trinket is 6, and a steel trinket is 3

using the well known winter acrobatics theorem (loll) , \(\frac{d}{dx}=\)\(nx^{n-1}\)

82170/99000=0.83

1-0.83=0.17

ans is 17

Use quadratic formula   a = 5     b= -11      c = 4

to find   x      11/10 +- sqrt 41 /10

1/a  + 1/b     =     10 /(11 + sqrt41)  + 10 / (11-sqrt41)

                   =    (110 + 110 )/( 121-41)

                    = 220 / 80

                    = 11/4                         Just as BB found !

then 1b + 1/ a = 2.75

We can solve this problem using Vieta's theorem. We want to find $\frac{1}{a} + \frac{1}{b}$, which simplifies to $\frac{b+a}{ab}$. Using Vieta's, we know that $a+b = -\frac{b}{a}$ and $ab = \frac{c}{a}$, where $a,b,c$ refer to a quadratic $ax^2+bx+c.$ Therefore, $a=5, b=-11, c=4. $ Using this, we find that $a+b = \frac{11}{5}$, and $ab = \frac{4}{5}$. Thus, we can find that $\frac{1}{a} + \frac{1}{b} = \boxed{\frac{11}{4}}.$

422     +     422 (.01)    + 422 ( .15)   =   422 ( 1.16) = $ ...........

7.74   *  1.15    *  .35  = ..........                     ( markdown 65 %   leaves 35% of the pre-markdown price)

We know that $\triangle JKL \sim \triangle JMN$ by $AA$ similarity ($\angle JKL \cong \angle JMN$ and $\angle JLK \cong \angle JNM$ because $KL \parallel MN$). Therefore,

$$\frac{JK}{JM} = \frac{JL}{JN},$$

$$\frac{36}{JM} = \frac{48}{60}, $$

$$JM = 45. $$

Given that $JM=45$, we see that $MK = JM - JK$, and $JK=36$ using the similarity ratio above. Thus, $MK = 45 - 36 = \boxed{9}.$

JM / JK   =JN/JL

JM / 36 = 60/48

JM =   36 * 60 / 48 = 45   units          (obviously not drawn to scale )       MK = JM - 36 = 9

SUm of triangles angles = 180

13x - 8      + 3x       +  7x+4 = 180

23x = 184

x =  8                                  13 (8) - 8 =                         3 (8) =                 7(8)+4 =          

ok so if we find the first term...

we calculate the absolute value first so it is 4.2 and then floor it, giving us 4

for the second term, it is -5 because that is what the floor of -4.2 is... the absolute value of -5 is 5

4 + 5 = 9

First, we can graph the parabola by graphing $y=(x-2)^2+1.$ The vertex of the parabola is $(2,1)$ and the $y$-intercept is where $x=0$. This turns out to be $y=(-2)^2+1 = 5$, so the $y$-intercept is $(0,5)$. Therefore, we can use this information to graph the parabola. We find the solution set of this by setting a test point, $(0,0).$ If this test point satisfies the inequality, then the region in which the test point is included is the solution set. I'll let you take it from here. 

(height of tree) / (shadow of tree) = (Andrea's height) / (Andrea's shadow)

Convert everything to inches beforehand.

(40 * 12)/(10 * 12) = x/15

x = 60 inches

If we have a polynomial with variable $x$, we can write the polynomial in factored form given its roots. Therefore, we get 

$$(x-(-2))(x-3)(x-4).$$ When we expand this out, we get $$x^3-5x^2-2x+24.$$ When we look at the other condition in the problem, we must have an $y$-intercept with a value of $24$. The $y$-intercept is where $x=0$, so we just look at the constant term in our polynomial and see that it is $24$. Thus, the least degree polynomial is $\boxed{x^3-5x^2-2x+24}.$

Because the lines are parallel, angle 1 = $\boxed{82}$.

similarly, angle 2= 180-a, and a=58, so angle 2 = $\boxed{122}$

because it's a parallelogram, x+x-26=180, or x = $\boxed{103}$, so x+26=$\boxed{129}$

hi someone already answered this heres the link

 https://web2.0calc.com/questions/rico-can-walk-3-miles-in-the-same-amount-of-time

1. wrote b incorrectly

2. y=4x+4

(30 x 9) = 270

270/5= 54

54+32= 86

F=86

plug c for 30, so F=270/5+32, so F=54+32=$\boxed{86}$

Because I cannot see it, I'm just assuming that another side is right, so 360-(145+65+90)=$\boxed{60^\circ}$

Let the cubic be in the form f(x) = ax^3 + bx^2 + cx + d.

We have four points and four variables, so plug them in:

(3, -4) ==> 27a + 9b + 3c + d = -4

(-1, 2) ==> a - b + c - d = 2

(0, 0) ==> d = 0

(7, 5) ==> 343a + 49b + 7c + d = 5

Use WolframAlpha or a similar program to solve for a, b, c, and d.

5 apples / 8 friends = $\boxed{\frac58}$

Force of friction for f_fz = u mg  = ma         ug = a   for A

    initial velocity 7.21

      vf= vi + a t       t is given as .05

       vf = 7.21 - .045(9.81)(.05)

          vf = 7.188     but velocity is only    3.81

              7.188 - 3.81 = 3.378 m/s  component of momentum   given up to B

                    so B should be moving initially at   (disregarding friction presently):

                       Using conservation of momentum for the system:

                           momentum A lost / mass B = velocityB                        3.378 (4.54) / 1.53  =10.023 m/s

Now include the friction component for .05 sec

For B    vf = vi - at

                 = 10.023 - .110(9.81)(.05) = 9.97 m/s   velocity of B after the objects separate              ( not positive about this answer)