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4 years ago

A carpenter has a board that is 10 feet long. He wants to make 6 table legs that are all the same length. What is the longest ea

ch leg can be?

Mathematics

1 answer:

Sphinxa [80]4 years ago

3 0

$You\ need\ to\ divide\ 10\ feet\ by\ amount\ of\ legs:\\\\10:6=1.66666\\\\We\ need\ to\ round\ this\ number\ down:\\\\1.6666=1.65\\\\The\ longest\ legs\ ca\ be\ 1.65\ ft.$

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rewrite the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as perfect squares

$(u+1/2)^2=9/4$

square root both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

<em>Find the solutions of the original equation</em>

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

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A ladder is leaning against a building.The distance from the bottom of the ladder to the building is 6ft shorter than the length

Akimi4 [234]

Answer:

15 ft

Step-by-step explanation:

Hi, the illustration for the problem is a right triangle with:

hypotenuse (C)= the length of the ladder = L

horizontal side(A) = distance from bottom of the ladder to the building = L - 6

vertical side(B) = distance from the top of the ladder to the bottom of the building = L - 3

So, we can use Pythagoras formula:

A2 +B2= C2

(L – 6 )² + (L-3)² = L²

L²-12L+36+L²-6L+9 =L²

L2 -18L+45 =0

APPLYING QUADRATIC FORMULA WE OBTAIN:

L =15 OR L=3

If L=3

Vertical side = L-3 = 0 (Length can´t be 0)

So L=15

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3 years ago