All Answers 358456 Answers

\(-\frac{1}{2}(7z+4)+\frac{1}{5}(5z-15)\)

First, let's use the distributive property

We get

\(-\frac{7}{2}z-2+z-3\)

I like using decimals, so this simplifies to

\(-3.5z-2+z-3\)

Now, let's just evaluate

\(-2.5z-5\)

That's the answer

Great answer MIRB14    

I am not good at presenting proofs properly but here is my logic.    

Let X be the centr of the circle.

Equal sectors cut of equal segments so the perpendicular heights from the centre of each segment chord to the circumference is going to be equal.

AX=BX=CX=DX=EX=FX       all equal radii

So

\(\triangle ABX \cong \triangle CDX \cong\triangle EFX\)

and all these triangles are isoscles with AB, CD, and EF respectively being the base sides.

The perpendicular bisector of the base sides will pass throught the centre of the circle. They are all altitudes off the base sides.

Since the altitudes are all the same, the circle that passes through these bisector points is concentric with the original circle.

Hence the centre of the incircle is also the centre of the original circle.

This definitely needs to be tidied up but do you understand what I am saying?

PLEASE HELP!!!!

3:

(a)     x - 1   =   \(\sqrt{2x+22}\)          and          The extraneous solution is  -3  .

(b)      bx + c  must be positive or zero because the square root of a negative number is imaginary.

2:      x + a  =  \(\sqrt{bx + c}\)

For  a , I pick  1 .  For  b , I pick  2 .  And they tell us that  x = 7 .

7 + 1  =  \(\sqrt{2(7) + c}\)        Now we just need to solve this for  c .

64  =  14 + c

c  =  50              So...

(a)      \(x + 1=\sqrt{2x+50}\)

----------

(b)      Square both sides to get...

(x + 1)²  =  2x + 50

x² + 2x + 1  =  2x + 50      Subtract  2x  from both sides and subtract  1  from both sides.

x²  =  49                            Take the  ± square root of both sides.

x = 7    or    x = -7

Test each potential solution.'

7 + 1  =  \(\sqrt{2(7)+50}\)     ?

8  =  \(\sqrt{64}\)      True.

-7 + 1  =  \(\sqrt{2(-7)+50}\)

-6  =  \(\sqrt{36}\)     False.

So  7  is the solution.  -7  is an extraneous solution.

----------

If  a = 2 ,  b = 2,  and  x = 7 , the equation is...

\(7+2=\sqrt{2(7)+c}\)       And  c  must equal...

c  =  67    So......

(c)      \(x+2=\sqrt{2x+67}\)

Yah, or maybe someone already answered it

1:

x + 2  =  \(\sqrt{3x+10}\)

                                           Square both sides of the equation.

(x + 2)²  =  3x + 10

(x + 2)(x + 2)  =  3x + 10

x² + 4x + 4  =  3x + 10

                                           Subtract  3x  from both sides and subtract  10  from both sides.

x² + x - 6  =  0

                                           Factor the left side.

(x + 3)(x - 2)  =  0

                                           Set each factor equal to zero and solve for  x .

x = -3    or    x = 2

Plug in  -3  into the original equation to see if it makes it true.

-3 + 2  =  \(\sqrt{3(-3)+10}\)      ?

-1  =  \(\sqrt{1}\)      ?

-1  =  1       False.

Plug in  2  into the original equation to see if it makes it true.

2 + 2  =  \(\sqrt{3(2)+10}\)      ?

4  =  \(\sqrt{16}\)      ?

4  =  4         True.

So....

(a)     2  is the solution.

(b)    -3  is the extraneous solution, because it does not make the given equation true.

Nice talk

The line  y + 6  =  -3x  has a y-intercept at  (0, -6) . This rules out the top two graphs.

The line  y = \(\frac13\)x + 4  is graphed correctly on both of the bottom two graphs.

The inequality says that  y  <   \(\frac13\)x + 4  , so the side below the line must be shaded.

That rules out the fourth graph. (It would have been right if the inequality was  y  >  \(\frac13\)x + 4 ).

So the third option is correct.  Here's a graph of the three inequalities from desmos:

First we can see that  one of the lines has the equation  y = x , and the shade is below this line,

so  y ≤ x  is one of the inequalities.

Then the dotted line has the equation  y = -2x - 2 , and the shade is above this line

so one of the inequalities must be  y > -2x - 2

The only option that has those two is the last one.  The last option is the correct one.

No. You have to balance it out.

3x+8=4x+3

subtract 3 on both sides

3x+8-3=4x+3-3

that gives

3x+5=4x

Then, this is the part where you messed up

We have to subtract 3x on both sides, not add 3x

that gives

3x+5-3x=4x-3x

Which we get

5=x

Usually, the varible is on the left side and the constant is on the right, so the answer would be

x=5

but 5=x is fine too

If the cost of 1 shoe =S, and:

The cost of 1 shirt =s, then:

4S =3s, then:

s =4S/3, but we have:

6S +3s =348. Sub s into this..............(1)

6S + 3(4S/3) =348

6S + 4S = 348

10S = 348

S=$34.80 - cost of 1 shoe, and:..............(2)

s =(4*34.80)/3

s=$46.40 - cost of 1 shirt. So that:..............(3)

4 shoes(2pairs)*$34.80 =3 shirts * $46.40

$139.20 = $139.20, and subbing (2) and (3) into 1 above, we have:

6*$34.80 + 3*$46.40 =348

$208.80 + $139.20 = $348.00, Therefore the cost of 25 shirts would be:

25 x $46.40 =$1,160.00

Given:    2 shoes = 3 shirts     and   6 shoes + 3 shirts = 348

Substitute:

3 shirts (3) + 3 shirts = 348

shirt = 29

then 25 shirts =   25 x 29 = 725.

(I should clarify that I ASSUMED  ' George bought 6 identical shoes'  means   '6 PAIRS' of shoes  in my calculation)

If you are aware of the way in which complex numbers are added multiplied and divided on an Argand diagram then yes it is possible to avoid much of the algebra.

Suppose that P and Q are points representing the complex numbers w and z, that O is the origin, then the complex number w+z will be at the point which is the fourth vertex of the parallelogram defined by OP and OQ. Because, in this case, w and z both lie on the unit circle, the parallelogram will be a rhombus and in that case its diagonal will bisect the angle POQ.

So, if arg(w) = theta and arg(z) = phi (and if, for convenience, phi > theta)

then arg(w + z) = theta + (phi - theta)/2 = (phi + theta)/2 ................... (1)

With multiplication, the moduli of the numbers are multiplied and the arguments added, that means that |wz| = 1

and arg(wz)= (phi + theta).

Applying the rule for addition again, (arg(1) = 0),  so arg(1 + wz) = 0 + (phi + theta - 0)/2 = (phi + theta)/2 ................... (2)

With division, one modulus is divided by the other and one argument subracted from the other (the reverse of the operation for multiplication), so, from (1) and (2), arg[(w + z)/(1 + wz)] = 0. That means that the number lies on the real axis and so is real.

If you are going to solve this problem algebraically though, involving the circular functions sine and cosine is the best, easiest, way to go.

Tiggsy

are you from ajman uni?

sin θ   =   opposite / hypotenuse

sin 60°   =   JK / 20

20 sin 60°   =   JK

JK   =   10√3

cos θ   =   adjacent / hypotenuse

cos 60°   =   JL / 20

20 cos 60°   =   JL

JL   =   10

area of JKL   =   (1/2) * JL * JK   =   (1/2) * 10 * 10√3   =   50√3    sq units

Yeah, I think it's because the post is too old. 

At least, I think that's what it is....

Whos He?

You could try reading your question before you post it. This is rubbish.

"When  is divided by 10, the remainder is . When  is divided by 13, the remainder is . What is  modulo 130, in terms of  and ?
(Your answer should be in the form , where  and  are replaced by nonnegative integers less than 130.)"

Most likely you are better to do it on a different calc but this is how it is done here.

\(15^\circ 14' 28''\)

(15+14/60+28/3600) = 15.2411111111111111   degrees