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Zarrin [17]
1 year ago
12

Three boards are placed end to end to make a walkway. The first board is 4 feet inches 10 long, the second board is 2 feet 11 in

ches long, and the third board is 4 feet inches 5 long. How long is the walkway?
Mathematics
2 answers:
sashaice [31]1 year ago
8 0

<em>Answer:</em>

<em>Answer:the answer is 10 feet and 16 inches</em>

<em>Answer:the answer is 10 feet and 16 inchesStep-by-step explanation:</em>

<em>b</em><em>y</em><em> </em><em>a</em><em>d</em><em>d</em><em>i</em><em>n</em><em>g</em><em> </em><em>a</em><em>l</em><em>l</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>feet</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>all</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>n</em><em>c</em><em>h</em><em>e</em><em>s</em><em> </em><em>w</em><em>e</em><em> </em><em>c</em><em>a</em><em>n</em><em> </em><em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>o</em><em>u</em><em>r</em><em> </em><em>answer</em>

  • <em>first</em><em> </em><em>we</em><em> </em><em>add</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>feet</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em>4 + 2 + 4 = 10</em>

  • <em>then</em><em> </em><em>we</em><em> </em><em>will</em><em> </em><em>add</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>inches</em>

<em>10 + 11 + 5 = 16</em>

  • <em>the</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em> </em><em>feet</em><em> </em><em>and</em><em> </em><em>1</em><em>6</em><em> </em><em>inches</em>

Strike441 [17]1 year ago
5 0

Answer:

12 ft 2 in.

Step-by-step explanation:

Conversion: 1 ft = 12 in.

Add the lengths of the boards.

4 ft 10 in. + 2 ft 11 in. + 4 ft 5 in. =

= 4 ft + 2 ft + 4 ft + 10 in. + 11 in. + 5 in.

= 10 ft + 26 in.

= 10 ft + 24 in. + 2 in.

Since 1 ft = 12 in., then 24 in. = 2 ft.

= 10 ft + 2 ft  + 2 in.

= 12 ft 2 in.

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Alex777 [14]

Answer:

x=1+t\\y=4-t\\z=4+2t

Step-by-step explanation:

A vector perpendicular to the plane ax+by+cz+d=0 is of the form (a,b,c).

So, a vector perpendicular to the plane x − y + 2z = 7 is (1,-1,2).

The parametric equations of a line through the point (x_0,y_0,z_0) and parallel to the vector (a,b,c) are as follows:

x=x_0+at\\y=y_0+bt\\z=z_0+ct

Put (x_0,y_0,z_0)=(1,4,4) and (a,b,c)=(1,-1,2)

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So, at point (-1,6,0)

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Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2

So, at point (0,5,2)

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So, at point (5,0,12)

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Hello there!
Goryan [66]

(x^2+3)^2 + 7x^2+21=-10

We want to convert this equation in terms of m, where m = x^2+3

While solving this problem, I instantly noticed the x^2+3 in the first term of the equation. Replace that with m.

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Do you see the "m" in here? Look again if you don't.

Now we can simplify the equation to m^2 + 7m=-10 and this cannot be simplified further.

I'm always happy to help someone who appreciates the help! :D

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