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3 years ago

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A carpenter used 1/3 able to finish of a box of nails while working on a birdhouse and was able to finish 1/5 of it. At this rat

e, how many boxes will he need to finish the entire birdhouse?

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer:

He will need 1 and 2/3 boxes

Step-by-step explanation:

1/5= 1/3 boxes

5/5= 1/3* 5 boxes

Hope this helps!

Let me know if it does

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Can someone help me with this problem please thank youu how do you do it ?

icang [17]

Answer:

A) x = 6 y = 6√3

Step-by-step explanation:

30-60-90 Triangle rule states that the hypothesis is 2x, the short leg is x, and the long leg is x√3. So, the short leg is 12/2 = 6, and the long leg is 6√3.

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3 years ago

Linda purchase a prepaid phone card for $30. Long distance calls cost $.17 a minute using this card when do use her card. Linda

balandron [24]

Answer: her call lasted for 44 minutes

Step-by-step explanation:

Linda purchase a prepaid phone card for $30. This means that the credit on her card is $30.

Long distance calls cost $0.17 each. Linda use her card only to make a long distance call. Assuming she made a total of x minutes of long distance calls, the cost would be $0.17x. The amount remaining on her card will be

30 - 0.17x

if the remaining credit on her card is $22.52, the number of minutes,x that the call lasted will be

22.52 = 30 - 0.17x

0.17x = 30 - 22.52

0.17x = 7.48

x = 7.48/0.17 = 44

7 0

3 years ago

I will give brainliest to whoever answers even if it’s just one part. please:)) i’ve been trying for so long

wariber [46]

Answer:

check your commments

Step-by-step explanation:

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3 years ago

Jacob is cutting yarn for a craft project. He needs pieces of yarn that are 12 inches long, but he only has a ruler marked with

lara [203]

Answer:

30.48 centimeters

Step-by-step explanation:

We know that:

2.54 cm = 1 inch

To find how many centimeters are in 12 inches, you should:

2.54 x 12 = 30.48

So Jacob’s pieces of yarn should be 30.48 centimeters long.

Hope this helps!

5 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:

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

<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