All Answers 356015 Answers

Hello Alan!

If brackets were set, it would look like this.
Without brackets 10 * 23 is not in the denominator.
10 ^ 23 is nowhere.
Thanks anyway! asinus :- ) 

(a-b+c-d)(a+b-c+d)

\(\begin{array}{|rcll|} \hline && (a-b+c-d)(a+b-c+d) \\ &=& [a-(b-c+d)][a+(b-c+d)] \qquad | \qquad (u-v)(u+v) = u^2-v^2 \\ &=& a^2-(b-c+d)^2 \\ &=& a^2-(b-c+d)(b-c+d) \\ &=& a^2-[(b-c)+d][(b-c)+d] \qquad | \qquad (u+v)(u+v) = u^2+2uv+v^2\\ &=& a^2-[ (b-c)^2+2(b-c)d+d^2 ] \\ &=& a^2-(b-c)^2-2(b-c)d-d^2 \\ &=& a^2-(b-c)^2-2bd+2cd-d^2 \qquad | \qquad (u-v)(u-v) = u^2-2uv+v^2\\ &=& a^2-(b^2-2bc+c^2)-2bd+2cd-d^2 \\ &=& a^2-b^2+2bc-c^2-2bd+2cd-d^2 \\ &=& a^2-b^2-c^2-d^2+2bc-2bd+2cd \\ \hline \end{array}\)

Please Help

Example: Found the team with index 2:

 \(\begin{array}{|lcll|} \hline 1. \text{Search: team index } \le 16 \qquad | \qquad \text{yes} \\ 2. \text{Search: team index } \le 8 \qquad | \qquad \text{yes} \\ 3. \text{Search: team index } \le 4 \qquad | \qquad \text{yes} \\ 4. \text{Search: team index } \le 2 \qquad | \qquad \text{yes} \\ 5. \text{Search: team index } \le 1 \qquad | \qquad \text{no} \\ \mathbf{6.} \text{Search: team index } = 2 \qquad | \qquad \text{yes and found} \\ \hline \end{array}\)

In the worst case 6 items. At most, 6.

-(x-2)=x-5-3x

\(-(x-2)=x-5-3x\)

\(-x +2 = x-5-3x\)

\(-x -x +3x =-5-2\)

\({\color{blue}x=-7}\)     !

I suspect this is meant to be \(\frac{2.72\times10^{22}}{6.02\times10^{23}}\rightarrow \frac{2.72\times10^{-1}}{6.02}\rightarrow0.4518...\times10^{-1}\rightarrow 0.04518...\)

.

sqrt(6^2+3^2) → sqrt(45) → sqrt(5*9) → sqrt(5*3^2) → 3sqrt(5)

.

The probability of a specific 5 bills containing errors is (0.6^5)*(0.4^5)

There are ncr(10,5) ways of choosing those 5 bills  ( ncr(10,5) = 10!/(5!*5!) )

so the overall probability is ncr(10,5)*(0.6^5)*(0.4^5)

.

sqrt(6^2+3^2) in surd form? ​

\(\sqrt{6^2+3^2 }\)

Klick ∑LaTeX

remove   x = {-b \pm \sqrt{b^2-4ac} \over 2a}

enter  \sqrt{6^2+3^2}

give  OK  ⇒

\({\color{blue}\sqrt{6^2+3^2}}\)    !

Thanks Alan!

\({\color{blue}\sqrt{6^2+3^2}=\sqrt{45}=\sqrt{5\times9}=3\times\sqrt{5}}\)

{\color{blue}\sqrt{6^2+3^2}=\sqrt{45}=\sqrt{5\times9}=3\times\sqrt{5}}

Greetings asinus :- )  !

\(\)

You should be able to find the item in no more than 6 tries.

First ask if the item is above half-way.  Whatever the answer you have reduced the search space so that you now only have 16 items to search.  Whichever 16 it is, choose the middle of them and ask if the item is above or below this position.  Now the answer will reduce the search space so you only have 8 items to search.  Continue in this way until there is only one possibility left.

What is the answer of this question: The algebraic expression of W more than 81 is...

W > 81

  !

how do I set the base of a log?

web2.0calc ⇒

cursor to "log"     (base 10)

below      "ln"       (base e)

below      "lg2"     (base 2)

  !

No thanks.

2.72*10^22/6.02*10*23

\(\frac {2.72\times 10^{22}}{6.02} \times 10\times 23={\color{blue}1.03920265781\times 10^{24}}\) 

  !

soh cah toa

Write a rule for a quadratic function with a graph that has x-intercepts (2,0) and (-6,0) and a maximum point of (-2,4)

quadratic function: \(y =ax^2+bx+c\)

1. Point (x=2, y=0):

\( \quad 0 = a\cdot (2^2)+b\cdot 2 + c \\ \Rightarrow 4a+2b+c=0\)

Maximum Point (\(x_{max} = -2, \ y_{max} = 4\)):

\(\begin{array}{|rcll|} \hline x_{max} &=& -\frac{b}{2a} \\ -2 &=& -\frac{b}{2a} \\ 2 &=& \frac{b}{2a} \\ 4a &=& b \\ \mathbf{ a } & \mathbf{=}& \mathbf{\frac{b}{4} }\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 4a+2b+c&=&0 \quad | \quad a=\frac{b}{4} \\ 4\cdot \frac{b}{4} +2b+c&=&0 \\ b+2b+c&=&0 \\ 3b+c&=&0 \\ 3b&=&-c \\ \mathbf{ b } & \mathbf{=}& \mathbf{-\frac{c}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a &=& \frac{b}{4} \quad | \quad b = -\frac{c}{3} \\ a &=& \frac{ -\frac{c}{3}}{4} \\ a &=& -\frac{c}{12} \\\\ y_{max} &=& a\cdot ( x_{max} ) ^2 + b\cdot x_{max} + c \quad | \quad a = -\frac{c}{12} \quad b = -\frac{c}{3} \\ y_{max} &=& -\frac{c}{12}\cdot ( x_{max} ) ^2 -\frac{c}{3}\cdot x_{max} + c \quad | \quad x_{max} = -2\\ y_{max} &=& -\frac{c}{12}\cdot ( -2 ) ^2 -\frac{c}{3}\cdot (-2) + c \quad | \quad y_{max} = 4\\ 4 &=& -\frac{1}{3}\cdot c +\frac{2}{3}\cdot c + c \\ 4 &=& \frac{4}{3}\cdot c \\ \mathbf{ c } & \mathbf{=}& \mathbf{3} \\\\ b&=&-\frac{c}{3}\\ \mathbf{ b } & \mathbf{=}& \mathbf{-1} \\\\ a&=&-\frac{b}{4}\\ \mathbf{ a } & \mathbf{=}& \mathbf{-\frac{1}{4}} \\ \hline \end{array}\)

quadratic function:

\(\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \\ \mathbf{ y }&\mathbf{=}& \mathbf{-\frac{1}{4}x^2-x+3} \\ \hline \end{array}\)

One plus also, (Assuming also is a variable), would be impossible to add. The simplest the answer could be is 1 + also.

If you doubt me, heres a great calculator you can use.

web2.0calc.com

Just divide:

13.40 / .20 =67 times that Sally-Anne had to wash and dry the dishes.

Expand the following:
(a-b+c-d) (a+b-c+d)

(a-b+c-d) (a+b-c+d) = a (a-b+c-d)+b (a-b+c-d)-c (a-b+c-d)+(a-b+c-d) d:
a (a-b+c-d)+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

a (a-b+c-d) = a a+a (-b)+a c+a (-d):
a a-a b+a c-a d+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

a a = a^2:
a^2-a b+a c-a d+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

b (a-b+c-d) = b a+b (-b)+b c+b (-d):
a^2-a b+a c-a d+b a-b b+b c-b d-c (a-b+c-d)+d (a-b+c-d)

b (-b) = -b^2:
a^2-a b+a c-a d+b a+-b^2+b c-b d-c (a-b+c-d)+d (a-b+c-d)

-c (a-b+c-d) = -(c a)-c (-b)-c c-c (-d):
a^2-a b+a c-a d+b a-b^2+b c-b d+-(c a)-(-c b)-c c-(-c d)+d (a-b+c-d)

-(-1) = 1:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c c-(-c d)+d (a-b+c-d)

-c c = -c^2:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b+-c^2-(-c d)+d (a-b+c-d)

-(-1) = 1:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d (a-b+c-d)

d (a-b+c-d) = d a+d (-b)+d c+d (-d):
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c-d d

d (-d) = -d^2:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c+-d^2

Grouping like terms, a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c-d^2 = -d^2+(c d+c d)+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b):
-d^2+(c d+c d)+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

c d+c d = 2 (c d):
-d^2+2 c d+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

-(b d)-b d = -2 (b d):
-d^2+2 c d+-2 b d-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

b c+b c = 2 (b c):
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

a d-a d = 0:
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a c-a c)+(a b-a b)

a c-a c = 0:
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a b-a b)

a b-a b = 0:
Answer: |-d^2 + 2cd - 2bd - c^2 + 2bc - b^2 + a^2

                                                      20 000  =  17 000(1 + r)⁵

Flip sides:                         17 000(1 + r)⁵  =  20 000

Divide both sides by 17 000:     (1 + r)⁵  =  20 000 / 17 000

Find the fifth root of both sides:     1 + r  =  (20 000 / 17 000)^1/5

Subtract 1:                                             r  =  (20 000 / 17 000)^1/5 - 1

Calculator time!

$20,000=$17,000(1+r)^5 divide both sides by 17,000

1.17647=(1+r)^5 take the 5th root of both sides

1+r =1.033 subtract 1 from both sides

r =0.033 x 100

r=3.3% 

Use the  10^x  key, entering 4.7893 for x.

nCr will ALWAYS give a positive integer answer!

\(5C2=\frac{5!}{2!3!}=10\)

\(5!=1*2*3*4*5\)

etc

Graph:  f(x)  =  -1/(x - 3) + 1

This becomes a hyperbola with asymptotes at:  x = 3  and  y = 1.

Since the coefficient of the x-term is negative, it will graph in the upper-left and lower-right portions of the plane separated by the asymptotes.

As the values for x get very large, the y-values approach 1 from below.

As the values for x get very small (large negative), the y-values approach 1 from above.

As the values for x approach 3 from the left, the values for y go to positive infinity.

As the values for x approach 3 from the right, the values for y go to negative infinity.

y  =  (1/3)x - 1

To find the x-intercept, replace y with 0 and solve:  0  =  (1/3)x - 1

                                                                                       1  =  (1/3)x

                                                                                       3  =  x      --->      (3, 0)

To find the y-intercept, replace x with 0 and solve:  y  =  (1/3)(0) - 1

                                                                                       y  =  -1      --->     (0, -1)