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what is 152.498395 corrected to 2 decimal places

what is 152.49 | 8395 corrected to 2 decimal places

9 is in the second decimal place so I have put a bar after it the next number ,8, is used for rounding and the numners after that are discarded.

\(152.49 | 8\not{3}\not{9}\not{5}\)

Since 8 is a big number (5,6,7,8,and 9 are big numbers) the 9 gets rounded up.

you cannot round up 9 so you will need to round the 49 up to 50

so the answer is

152.50

To increase an amount by 3.2% what single multiplier would you have to use?

Well you want 100% + 3.2% = 103.2% 

103.2% = 103.2/100  =   1.032

The following might help:

Use the cosine rule to get the length of the resultant:  \(c^2=a^2+b^2-2ab\cos\theta\)

y = xln(x)^2-3x^-7

\(y = x^{ln(x)^2}-3x-7\\~\\do \;you \;mean\\ y = x^{(ln(x))^2}-3x^{-7}\\or\\ y = x^{ln(x^2)}-3x^{-7}\\\)​

Simply plug it in the calculator

2.2+1 = 3.2    or  16/5

ขออภัย แต่นี้ไม่ได้แปลอย่างถูกต้อง

I do not understand this notation.  Would you like to expand on it ?

Thanks Chris :)

Oh yea, I overlooked the years.  

Ok I'll bite,

Who's there ?

Rate of change =

[ 12250 - 12000] / [ 2012 - 2005]  =

250 / 7   = about 36 students were added per year

+7 - 4 - 2 - 10 + 5 + 4  =

7 - 6 - 10 + 9  =

1 - 10 + 9   =

-9 + 9  =

0   !!!!

In 2005, there were 12,000 students in a high school. In 2012, there were 12,250. What is the rate of change in the number of students?

The change is 250 students in the time period.   The time period should have been specified in the question.

The original was 12000 students

the rate of change was    250/1200   in the time period  

or    

\(250/1200*100\% =  250/12\% = 20.8\dot3\%\)

How do you know that sin45º is √2/2?

Start by drawing a right angled isosceles triangle.  Let the legs be one unit long.

Use the Pythagorean theorem to get the length of the hypotenuse.

\(sin45^0=\frac{opp}{hyp}=\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{2}\)

I have never memorised any of the 30,60,45 degree trig ratios, I have only memorised the triangles that are used to work them out.  That is my preference, not everyone's  :)

We can solve this with a simple proportion

your height                                  flagpole height    

------------------------------     =        --------------------------------------------------  

length of your shadow                length of shadow cast by the flagpole

6/ 4    =     flagpole height /  12     .......  multiply both sides by 12

12 [6/4]    = flagpole height

72/4   = flagpole height

18 ft   = flagpole height

Here's the expansion :

x^8 + 4x^5 + 6x^2 + 4/x+ 1/x^4

No constant term exists  !!!

[ 7 + 1/5]  - [6 + 2/5]    

Note that 7 + 1/5   =   6 + 1  + 1/5    =   6 + 5/5 + 1/5   =  6 + 6/5

So  now....it should be easy

[ 6 + 6/5]  - [6 + 2/5]   =  

[ 6 - 6 ] + [ 6/5 - 2/5]  =

4/5

a.   3x   - 2y  = 14

     -2x  + 2y  = -10         just add the equations   and we have :

x  =  4         to find y......just substitute for x in either of the equations

3(4)  - 2y   = 14  

12  - 2y  =14        subtract 12 from both sides

-2y  = 2              divide both sides by -2

y = -1  

So......the solution is :  (x, y)   = ( 4, -1)

b.   y = 5x + 3

     -2x - 4y  = 10

Substitute the first equation for y into the second equation

-2x - 4[5x + 3]  = 10       simplify

-2x - 20x - 12  = 10

-22x  - 12  = 10      add 12 to both sides

-22x  = 22       divide both sides by -22

x = -1

And using y = 5x + 3,   we have that

y = 5(-1) + 3   = -2

So........the solution is   (x, y)   = ( -1, -2)

When f(x) = x^3 + 3x^2 + ax + b is divided by x + 2 and x - 1, the remainders are -4 and 8, respectively. Find the values of a and b.

By the Remainder Theorem, f(-2)  = -4  and f(1)  = 8........so.....we can set up this system :

[-2]^3 + 3[-2]^2 + a(-2) + b   = -4

[1]^3  + 3[1]^2 + a (1) + b  = 8            simplify  these

-8 + 12 - 2a + b   = -4

1 + 3   + a  + b  = 8            simplify again

-2a + b  = -8

  a + b  =    4      (1)

Multiply     (1)   through by - 1

-2a + b  = -8

 -a  - b   = -4         add the equations

-3a  = -12    →   a =  4

And.......using (1)   to find b

4 + b = 4      →   b =  0

So.....the original function is

f(x) =  x^3 + 3x^2 + 4x

Here's the graph :   https://www.desmos.com/calculator/xi4bdnrstr

a. An expression for the area is :  L * W  =   (8 + 2x) ( 10 + x + [3x - 1])   = (8 + 2x) (9 + 4x)

b. If the border is one inch, then x = 1, and we have :

  [8 + 2(1)] * [ 9 + 4(1) ]  = [ 10 ] * [13]  = 130 in^2

c. We need to solve this :

(8 + 2x) ( 9 + 4x)  = 165      simplify

8x^2 + 50x + 72  = 165      subtract 165 from both sides

8x^2 + 50x - 93  = 0    factor

[4x + 31] [ 2x -3 ]  = 0

Setting the first factor to 0  produces a negattive value for x

Setting the second factor  to 0, we have that

2x - 3  = 0     add 3 ro both sdes

2x  = 3     divide both sides by 2

x = 3/2 inches  = 1.5 inches

2.4*(-8.3) = -19.92

I think you meant this :

(1-costheta)/(1+costheta)=tan^2(theta/2)

tan ^2 (theta/2)  =

sin^2 (theta/2) / cos^2 (theta/2)  =

 (±√  ( [ 1 - cos theta] / 2 ) )^2 /  (±√ ( [ 1 + cos theta] / 2 ) )^2

( [ 1 - cos theta] / 2 )  /  ( [ 1 + cos theta] / 2)  =

[1 - cos theta] / [1 + cos theta]  =  the left hand side

a.  Taking the height of the water's surface  as 0, we have

-16t^2 + 6t + 12  = 0      divide through by -2

8t^2 - 3t - 6  = 0            this will not factor.....using the quadratic formula, we get that t =

about 1.07  seconds

b. The max height above the pool   is given by  c - [ b^2 ] / [4a]   =  12 - [36]/ [ 4(-16)]  =

12 + 36/64   =

12 + 9/16 ft  =

12.5625 ft