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3 years ago

What square root equals 1.1?

Mathematics

2 answers:

vovangra [49]3 years ago

8 0

The square root that equals 1.1 is 7.

Ilya [14]3 years ago

6 0

Whats the opposite of square root? Square (exponent)
So do 1.1 squared.
1.21 is the answer.
jheez eighth grade is easy

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Please help me llxl+2l=2

miskamm [114]

$||x|+2|=2$

if x≥0 then |x|=x
if x<0 then |x|=-x

1) for x≥0

$|x+2|=2 \\
x+2=2 \ \lor \ x+2=-2 \\
x=0 \in D \ \lor \ x=-4 \notin D$

2) for x<0

$|-x+2|=2 \\
-x+2=2 \ \lor \ -x+2=-2 \\
-x=0 \ \lor \ -x=-4 \\
x=0 \notin D \ \lor \ x=4 \notin D$

The answer is:
$x=0$

3 0

3 years ago

Read 2 more answers

The ages of 3 siblings combined is 27.the oldest is twice the age of the youngest.the middle child is 3 years older than the you

Olenka [21]

I ended up with the ages of 12, 3, and 2
if u multiply 12 by 2 u end up with 24 then u add 3 and get 27
but I think i have i worked out wrong

8 0

4 years ago

A catering company makes fruit salad using only 4 ingredients. They use 4 apples for every mango, 2 pears for every apple, and 1

Advocard [28]

Well, if they used 8 pineapples, that means they used 8 mangos, which means they used 32 apples, and 64 pears
8 pineapples = 8 mangoes
8 mangoes x 4 apples = 32 apples
32 apples x 2 pears = 64 pears

7 0

3 years ago

Which shows the following in order from least to greatest: 85%, 15/18, 0.84 PLEASE HELP

Marrrta [24]

Answer:

15/18, 0.84, 85%

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

6 0

3 years ago

Read 2 more answers

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$