All Answers 358087 Answers

Thank you very much, Will. 

Thank you very much, Alan. 

It is given that they are a junior, so we will only look at the junior row.

\(\frac{16}{16+2+8} \)

\(=0.615\)

There are 26 juniors of whom 16 preferred morning classes, so the probability is 16/26 = 0.615

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Use atan (Press 2nd then atan on the calculator on the home page here)

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As follows:

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Using the binomial theorem how would you simplify (sqrt(x) +5)^3?..........sqrt= squareroot

\(\begin{array}{lcl} (\sqrt{x} +5)^3 &=& \binom30 \cdot (\sqrt{x})^3 \cdot 5^0\\ && + \binom31 \cdot (\sqrt{x})^2 \cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1 \cdot 5^2 \\ && + \binom33 \cdot (\sqrt{x})^0 \cdot 5^3 \\\\ &=& \binom30 \cdot (\sqrt{x})^3 \cdot 1 \\ && + \binom31 \cdot (\sqrt{x})^2 \cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1 \cdot 5^2 \\ && + \binom33 \cdot 1 \cdot 5^3 \\\\ &=& \binom30 \cdot (\sqrt{x})^3 \\ && + \binom31 \cdot (\sqrt{x})^2\cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1\cdot 5^2 \\ && + \binom33 \cdot 5^3 \\\\ &=& \binom30 \cdot x^{ \frac{3}{2} } \\ && + \binom31 \cdot x^{ \frac{2}{2} } \cdot 5^1 \\ && + \binom32 \cdot x^{ \frac{1}{2} } \cdot 5^2 \\ && + \binom33 \cdot 5^3 \\\\ &=& \binom30 \cdot x^{ \frac{3}{2} } + \binom31 \cdot x \cdot 5^1 + \binom32 \cdot x^{ \frac{1}{2} } \cdot 5^2 + \binom33 \cdot 5^3 \end{array} \)

\(\begin{array}{lcl} \binom30 = \binom33 = 1 \\ \binom31 = \binom32 = 3 \end{array}\)

\(\begin{array}{lcl} (\sqrt{x} +5)^3 &=& 1 \cdot x^{ \frac{3}{2} } + 3 \cdot x \cdot 5^1 + 3 \cdot x^{ \frac{1}{2} } \cdot 5^2 + 1 \cdot 5^3 \\\\ &=& x^{ \frac{3}{2} } + 15 \cdot x + 75 \cdot x^{ \frac{1}{2} } + 125 \\\\ \mathbf{(\sqrt{x} +5)^3} & \mathbf{=}& \mathbf{ x^{1.5} + 15 \cdot x + 75 \cdot x^{0.5} + 125 } \end{array}\)

$1320 tax was paid in 1999. 10% less was paid in 2000.

a. calculate how much was paid in 2000.

$1320 was 10% more than what was paid in 1998.

b. calculate the tax that was paid in 1998

a-$1,320 x .90 =$1,188 tax paid in 2000

b-$1,320/1.10=$1,200 tax paid in 1998

You can expand as follows:

Expand the following:
(sqrt(x)+5)^3

(5+sqrt(x))^3  =  sum_(k=0)^3 binomial(3, k) (sqrt(x))^(3-k) 5^k  =  binomial(3, 0) (sqrt(x))^3 5^0+binomial(3, 1) (sqrt(x))^2 5^1+binomial(3, 2) (sqrt(x))^1 5^2+binomial(3, 3) (sqrt(x))^0 5^3:
binomial(3, 0) x^(3/2)+5 binomial(3, 1) x+25 binomial(3, 2) sqrt(x)+125 binomial(3, 3)

binomial(3, 0) = 1, binomial(3, 1) = 3, binomial(3, 2) = 3 and binomial(3, 3) = 1:
5^3+5^2×3 sqrt(x)+5×3 (sqrt(x))^2+(sqrt(x))^3

5^2 = 25:
5^3+25×3 sqrt(x)+5×3 (sqrt(x))^2+(sqrt(x))^3

Cancel exponents. (sqrt(x))^2 = x:
5^3+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5^3 = 5×5^2:
5×5^2+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5^2 = 25:
5×25+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5×25  =  125:
125+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

25×3  =  75:
125+75 sqrt(x)+5×3 x+(sqrt(x))^3

5×3  =  15:
125+75 sqrt(x)+15 x+(sqrt(x))^3

Multiply exponents. (sqrt(x))^3 = x^(3/2):
Answer: |  125+75 sqrt(x)+15 x+x^(3/2)

why are you asking random rude questions on a calculator site? This site is not only for a calculator...if you weren't ignorant maybe you would have figured that out. :)

The answer is D

By factorizing it becomes: (z^4+z^2+1)*(z+1)^2=0

13.33333333333333333333%

Z=0

638

the answer is 77,000

3+3+3+3=12

The answer is 12 mi

650-(6+5+5)=634

634 is the missing number

n=35042

Thanks guest,

I only just saw this :))

Lets see

\(CosA=\frac{1+sinA}{e^x}\\ CosA=\frac{1+(1-cos^2A)^{0.5}}{e^x}\\ e^xCosA=1+(1-cos^2A)^{0.5}\\ e^xCosA-1=(1-cos^2A)^{0.5}\\ e^{2x}Cos^2A-2e^xCosA+1=1-Cos^2A\\ e^{2x}Cos^2A+Cos^2A-2e^xCosA=0\\ (e^{2x}+1)Cos^2A-2e^xCosA=0\\ (e^{2x}+1)Cos^2A-2e^xCosA=0\\ cosA[(e^{2x}+1)CosA-2e^x]=0\\ cosA=0 \qquad or \qquad (e^{2x}+1)CosA-2e^x=0\\ cosA=0 \qquad or \qquad CosA=\frac{2e^x}{ e^{2x}+1 }\\\)

What is the square root of 81*4

\(\sqrt{81*4}=\sqrt{81}*\sqrt4= 9*4=36\)

1&((1&(0^1)|((1&1^1)^1))^1)|((1&1^1&(1&(0^1)|((1&1^1)^1)))^1)&0

1&((1&0)|((1&1)^1))^1)|((1&1&(1&(0)|((1&1)^1)))^1)&0

What are all the &s for ?

Hello Guest!

expand sin2x

sin 2x = 2 sinx cosx

greeting asinus :- ) !