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3 years ago

15

In this fulcrum, the weights are perfectly balanced. How far must the fulcrum be located from the 40 lb. weight if the bar is 9

feet long?

2 answers:

Elis [28]3 years ago

5 0

Answer:

$x=5\ ft$

Step-by-step explanation:

we know that

The bar is $9$ feet long

so

$x+(x-1)=9$

Solve for x

$2x-1=9$

$2x=10$

$x=5\ ft$

Verify

$40x=50(x-1)$

substitute the value of x

$40(5)=50(5-1)$

$40(5)=50(4)$

$200=200$ -----> the weights are perfectly balanced

creativ13 [48]3 years ago

4 0

So...we can write eqn as

40x = 50(x-1)
so. -10x=-50
so. x=5

therefore .... the distance of 40 lb will be 5 ft ...answer !!!

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Which expression is equivalent to |a|≤3 ?

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Explanation:

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Therefore, $|a| \le 3$ turns into $-3 \le a \le 3$ which breaks down further to $a \ge -3 \ \text{ and } \ a \le 3$

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F(t) = -(t-2) (t-15)

Goryan [66]

Answer:

smaller (t) = 2 and this in coordinates form (2,0)

larger (t) = 15 and this in coordinates form (15,0)

the vertex of the parabola = $(\frac{17}{2} ,\frac{169}{4} )$ or in decimals (8.5, 42.25)

Step-by-step explanation:

For the zeros of the function take each bracket and make it equal to zero.

SMALLER (t):

(t-2)=0

(add 2 for both sides)

t=2

LARGER (t):

(t-15)=0

(add 15 for both sides)

t=15

For the vertex of the parabola you do:

step1: expand the brackets:

f(t)=-(t-2)(t-15)

f(t)= $-t^{2} +17t-30$

step2: define a,b and c using the expression $ax^{2} +bx+c$ :

a= -1 (the coefficient of $t^{2}$ )

b= 17 (the coefficient of t)

c= -30 (the single number without a letter)

step3: sub the values in the formula $-\frac{b}{2a}$ to find the x coordinate of the vertex :

$-\frac{b}{2a}$

= $- \frac{17}{2*-1}$

= $\frac{17}{2}$ or in decimals 8.5

step4: sub the value of t (the x-coordinate) in equation f(t):

f(t)= $-t^{2} +17t-30$

f ( $\frac{17}{2}$ ) = - $(\frac{17}{2} )^{2}$ + 17× $\frac{17}{2}$ - 30

= $-\frac{289}{4}$ + $\frac{289}{2}$ -30

= $\frac{169}{4}$ or in decimals 42.25

(THIS IS A PICTURE OF THE GRAPH↓)

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