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Let's call the number x. Twice the number means 2x, and half the number means x/2. We Multiply the whole thing by x, and that thing in total equals x.

Putting this into equation form...

x(2x+x/2)=x

2x^2+x^2/2=x

5x^2/2=x

5x^2=2x

5x=2

x=2/5

Therefore, our mystery number is 2/5.

You are very welcome!

:P

I went to Symbolab for this one:

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%5Cinfty%20%7D%5Cleft(%5Cfrac%7B%5Cleft(2x-1%5Cright)%7D%7B%5Cleft(x%5Cright)%5Cleft(x%2B1%5Cright)%5Cleft(x%2B2%5Cright)%7D%5Cright)

Set it so that \(x^2-3x+7=3x-2\) .

Now bring over to the left side: \(x^2-6x+9=0\)

We can find out that \(x=3\).

Now plug it in to \(y=3x-2 => y=3(3)-2 => y=7\)

Therefore, the y-coordinate is 7.

You are very welcome!

:P

Ricardo will have to pay 10% of the increase   MORE...

Increase = 700-550 = 150

10% x 150 = $15 more per month

First one

3x^3 = 0    implies that the zero  =  0

This zero has a multiplicity of 3  (the power on x)

With an odd multiplicity.....the graph goes through the x axis

(x - 1)^2 = 0    implies that the zero =  1

Multiplicity = 2     (same as the exponent)

With an even multiplicity....the graph just touches the x axis

(x - 8) = 0     implies that the zero =  8

The implied exponent is 1....so the multiplicity is 1

Odd multiplicity = goes through the x axis

Second one......using what we know

-x =0 implies the zero = 0

Multiplicity = 1

Goes through the x axis

5x+ 3 = 0

5x = - 3

x = -3/5      this is the zero

Multiplicity = 3     [ the exponent on (5x + 3) ]

Goes through the x axis

x - 2 =  0     implies that the zero = 2

Multiplicity  = 8    [ the exponent on ( x - 2) ]

Even multiplicity......just touches the x axis

Nice, Melody.....you have laid this out so it's easy to see!!!

Regardless of whether  is odd or even, we have 5 choices for : if  is odd then  must be one of the 5 even digits, otherwise if  is even then  must be one of the 5 odd digits. Similarly, we then have 5 choices for , 5 choices for , and so on.

Since  can be any of the 10 digits, the answer is 

\(f=x^2+2xy+3y^2-6x-2y\\ \nabla f=(2x+2y-6,2x+6y-2)\\ \nabla f = 0 \Rightarrow\\ 2x+2y=6\\ 2x+6y=2\\ 4y=-4\\ y=-1,~x=4\)

\(\text{This is a stationary point. We don't know it's a minimum.}\\ \text{We have to apply the 2nd derivative test in 2D}\\ f_{xx}=2\\ f_{yy}= 6\\ f_{xy}=2\\ \left|\begin{pmatrix}2 &2\\2&6\end{pmatrix}\right|= 8 > 0 \text{ and } f_{xx}=2>0\\ \text{Thus we have a minimum}\)

y=x^2−3x+7

y=3x−2

Set yhe y's equal and we have that

x^2 - 3x + 7  =   3x - 2          subtract 3x / add 2   from/to both sides

x^2 - 6x + 9   =  0        factor

(x - 3)^2 = 0        which implies that x = 3

So.......plug 3 into either function  and we get that   y = 3(3) - 2   =  7    

thanks

y = -2x + 1

y = -2x^2 + 1

Set the y's equal

-2x^2 + 1 = - 2x + 1           subtract 1 from both sides

-2x^2 = - 2x             multiply both sides by -1

2x^2 = 2x       divide both sides by 2

x^2 = x     subtract x from both sides

x^2 - x = 0       factor

x ( x - 1) = 0

Setting each factor to 0  and solving for x produces two  solutions at

x = 0      and   x =  1

In one hour Plane A travels  290 (1)  =  290 miles

Assuming that Plane B has a constant rate.....it travels   540 / 2 =   270 miles in one hour

So.....C is correct

sumfor(k, 1, 5, 3*(-2*k+5)) = (9, 3, -3, -9, -15) = -15

10.   f(2)   =  f ( √( x + 1] )    implies that

√ [ x + 1 ]  =  2   square both sides

x + 1 =  4

x = 3

So  

f(2)  =  1/3

\(\text{There are a few ways to go about this. Lagrange multipliers is one}\\ f = x^2 + y^2\\ g = (x^2 - xy+y^2 - 3)\\ \text{Solve }\nabla(f+\lambda g)=0\)

\(\text{This results in 3 equations}\\ 2x + \lambda(2x-y)=0\\ 2y+\lambda(2y-x) = 0\\ x^2 - xy +y^2 = 3\)

\(\text{This system has 4 solutions}\\ (x,y) = \pm(1,-1)\\ (x,y) = \pm (\sqrt{3},\sqrt{3})\\ \text{resulting in min and max of }x^2 + y^2 \text{ being }2 \text{ and } 6\\ m+M=2+6 = 8\)

I leave the gritty details to you.

Answered here :  https://web2.0calc.com/questions/which-answer-describes-the-function-f-x-2x-3-x-2_2

https://web2.0calc.com/questions/which-answer-describes-the-function-f-x-2x-3-x-2_2

WHY DIDN'T YOU CHECK YOUR PREVIOUS POSTING OF THIS QUESTION????????

Let the integers be (in ascending order)  a, b, c, d , e

Then we have these two possible systems

a + b = 16                   a + b  =  16

a + c = 20                   a + c   = 20

d + e = 37                   d + e   = 37

c + e = 34                   c + e =   34

a + d = 22                   b + c =   22

In the first system we have that

e - a = 14  and

e - a = 15

impossible

In the second system we have that

c = 22 - b

And

a + b = 16     (1) 

a + (22 - b) = 20  ⇒   a - b = - 2   (2)

Adding (1) and (2)  we have that  2a = 14 ⇒   a = 7

So

b = 9

c = 22 - 9  =  13

e = 34 - c = 34 - 13 =  21

d = 37 - e   =  37 - 21   = 16

So....the five integers are

7, 9, 13, 16, 21

\(\text{Let the numbers be }a,b,c,d,e \text{ listed in increasing order}\\ \text{You can get 4 equations straight off}\\ a+b = 16\\ a+c = 20\\ c+e=30\\ d+e=37\)

\(\text{solving this system gets you }\\ (a,b,c,d,e) = (a, 16 - a, 20 - a, 23 - a, 14 + a)\\ \text{You can then create the sums of pairs from this}\\ a+b = 16\\ a+c=20\\ a+d=23\\ a+e = 14+2a\\ b+c=36-2a\\ \text{etc.}\\ \text{plug in the handful of values for }a \text{ that are possible, }1-15\\ \text{sort the results and compare it with the given values}\\ \text{It turns out }a=7 \text{ is the winner, resulting in }\\ (a,b,c,d,e) = (7, 9, 13, 16, 21)\)

\(\left( \begin{array}{cc} \{a,b\} & 16 \\ \{a,c\} & 20 \\ \{b,c\} & 22 \\ \{a,d\} & 23 \\ \{b,d\} & 25 \\ \{a,e\} & 28 \\ \{c,d\} & 29 \\ \{b,e\} & 30 \\ \{c,e\} & 34 \\ \{d,e\} & 37 \\ \end{array} \right)\)

Sorry for the late response but I understand now.  I didn't even see the product rule between t^2 and sin(t).

For a kite ....there are two sets of sides that re equal length....

   so side a  + side b = 44

a = 3b -4

a + b = 44

3b-4 +b = 44

4b-4 =44

4b=48

b= 12            a = 3b-4 = 32

thanx Heureka......appreciated!!

As soon as I drew it myself the answers (or how to find them) become obvious.

ABCD    and     ABST   are both cyclic quadrilaterals

Opposite angles in cyclic quads are always supplementary. 

You can probably do it yourself now :)

Circles

Give with reasons ATS and TDS in terms of x...

In a chordal quadrilateral: Opposite angles are supplementary.

\(ATS = TDS = 180^\circ - x\)