All Answers 356004 Answers

The sum of the coordinates is 15/2.

The minimum value is 3/4.

$\mathbf{a}\bullet( \mathbf{b} + \mathbf{c}) = 7$

$\mathbf{b}\bullet(\mathbf{a} +\mathbf{b} + \mathbf{c}) = 8$

$(\mathbf{a} + 2\mathbf{b})\bullet(-3\mathbf{a} + 3\mathbf{c}) = 10$

By quadratic optimization, the smallest distance between the origin and the graph is $\boxed{2 \sqrt{2}}$.

2xº + 5xº + (x+40)º = 180º

sight distance= 99 / cos(15º)

https://web2.0calc.com/questions/geometry-help-asap-pls_1#r2

n=100; a=((1+2#5)/2)^n + ((1-2#5)/2)^n

L(100) ==792,070,839,848,372,253,127 mod 9 ==7 - the remainder

What is mn?

Hello Guest!

\(2x^2 - 6x + 5 = -4\\ x^2-3x+4.5=0\\ x=1.5\pm \sqrt{2.25-4.5}\\ x=1.5\pm \sqrt{-2.25}\\ x=1.5\pm 1.5i\)

\(m=1.5,\ n=1.5\\ \color{blue}mn=2.25\)

  !

a*b = 2/5.

YZ = 12 ³/₁₄      or     171/14      or       12.21428571  

Show your work, please. Your answer is way off!!!

1st step:       XZ / WX = WX / XY

2nd step:      YZ = XY - XZ

YZ = 22/5.

The smallest distance is 5*sqrt(6).

assuming single digit numbers are written without the leading zero

the mean is:

\(\frac{1+2+\cdots+9+1\cdot1+1\cdot2+\cdots1\cdot9+2\cdot1+2\cdot2+\cdots2\cdot9+\cdots\cdots9\cdot1+9\cdot2+\cdots9\cdot9}{100} \\=\frac{45+1(45)+2(45)+\cdots+9(45)}{100} \\=\frac{45\cdot46}{100} \\=\boxed{20.7}\)

suppose sin(a)+cos(b)=-0.2 while sin(b)+cos(a)=1.6 

What is the value ofsin (a+b)

\(sin(a)+cos(b)=-0.2 \qquad \qquad and \qquad\qquad sin(b)+cos(a)=1.6\\ (sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56\\ sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\ (1)+(2)\\ sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\ 1+1+2(sinacosb+sinbcosa)=3\\ 2(sinacosb+sinbcosa)=1\\ sin(a+b)=0.5\\\)

You need to check it.

LaTex:

sin(a)+cos(b)=-0.2 \qquad \qquad and  \qquad\qquad sin(b)+cos(a)=1.6\\

(sin(a)+cos(b))^2=(-0.2)^2 \qquad and \qquad \qquad (sin(b)+cos(a))^2=(1.6)^2\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \quad and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56\\
sin^2(a)+cos^2(b)+2sinacosb=0.04 \;\;(1)\;\; and \quad  sin^2(b)+cos^2(a)+2sinbcosa=2.56 \;\;(2)\\
(1)+(2)\\
sin^2(a)+cos^2(a)+sin^2(b)+cos^2(b)+2sinacosb+2sinbcosa=0.04+2.56\\
1+1+2(sinacosb+sinbcosa)=3\\
2(sinacosb+sinbcosa)=1\\
sin(a+b)=0.5\\

How about 3465

Two numbers, if the first one increases by 1, and the second one decreases by 1, then their product increases by 2020. If the first number decreases by 7, and the second one increases by 7, what value does the product decrease?

let the numbers be x and y   and let the product decrease by k

\((x+1)(y-1)=xy+2020\\ xy-x+y-1=xy+2020\\ -x+y-1=2020\\ y-x=2021\qquad (1) \)
 

\((x-7)(y+7)=xy - k\\ xy+7x-7y-49=xy - k\\ 7x-7y-49= - k\\ 7(x-y)-49=-k\\ k=49-7(x-y)\\ k=49-7*-2021\\ k=49+7*2021\\ k=14196\)          

it is up to the asker to check that the answer is correct.

atan stands for arctan     and     arctan is the same as inverse tan.

so

tan^(-1)

or

atan      

or

you can use the  x²  button in the ribbon and have    tan^-1

or

you can use Latex. Open the Latex box in the ribbon and type  tan^{-1}   and you get      \(tan^{-1}\)

A positive integer k is called a "lucky" number if the units digit of k^6 is 4. How many lucky numbers smaller than 1000 are there?

For this to happen the units value of k must be 2 or 8
so, 2,8,12,18, etc

1000/10 *2 = 200

there will be 200 lucky values of k

\(g^4 + 12g^2 + 9 - 6g^2 + 7\\ [g^4 + 12g^2+(\frac{12}{2})^2 ]+ 9 - 6g^2 + 7-(\frac{12}{2})^2\)

You can take it from there.

reposted here:

This is a repost of this one