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Pachacha [2.7K]

2 years ago

11

Sam built a fence that was 54 feet long she built the same legnth of fence each day for 6 days how many inches of fence did Sam

build each day

2 answers:

natima [27]2 years ago

7 0

Answer:

648 inches each day he built

Step-by-step explanation:

There are 12 inches in a foot. So 12 times 54= 648

FromTheMoon [43]2 years ago

4 0

Answer:

9 inches a day

Step-by-step explanation:

9 multiplied by 6 is 54

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A book sold 33,200 copies in its first month. Suppose this represents 7.4% of the number of copies sold to date. How many copies

Marta_Voda [28]

Answer: 448648

Step-by-step explanation:

33,200 copies represent 7.4% of the total amount

-----------------------------------------------------

#1 33,200 ÷ 7.4%

- this is because how we get 33,200 by calculation is use the total amount and times 7.4% to get 33,200. So we are just working backwards. Since it was times before, then we divide after.

33,200÷7.4%

=33,200÷0.074

≈448648 (you can't round up when the real amount is less the exact amount)

6 0

2 years ago

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Solving one step equations 1/4 a = 3

Pie

Answer:

a = 12

Step-by-step explanation:

$\frac{1}{4} a = 3$

Convert element into fraction : 3= 3/1

$\frac{1a}{4} = \frac{3}{1}$

Cross multiply

$1a \times 1 = 4 \times 3 \\ a = 12$

6 0

3 years ago

4/5 to 3/5 as a tenth of a percent

Rudik [331]

12/25 that will be the answer so its the percent time always helps.

5 0

2 years ago

What is the rule for the following pattern? 2, 6, 8, 24, 26

dimulka [17.4K]

Answer:

Always multiply the first number with 3 and then add two to the next number. And so on..

Step-by-step explanation:

6 0

3 years ago

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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:

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

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

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

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

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

$\overline{y} = 4$

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

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!

<u>3) Find the squared vertical distances between this line and the three points.</u>

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.

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

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

<u>4) Find the value of m that makes R minimum.</u>

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)

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

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

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

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

5 0

3 years ago