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3 years ago

6

Double the sum of a double the number and quotient of five and two

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

The required expression is $2(2x+\frac{5}{2})$

Step-by-step explanation:

Let the number be x

We are supposed to write the algebraic expression of given statement

Double the number = 2x

Quotient of five and two = $\frac{5}{2}$

Now we are given that Double the sum of a double the number and quotient of five and two

So, Sum of a double the number and quotient of five and two = $2x+\frac{5}{2}$

So, Double the sum of a double the number and quotient of five and two = $2(2x+\frac{5}{2})$

Hence The required expression is $2(2x+\frac{5}{2})$

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A toy rocket is launched from a platform that is 48 ft high. The rocket height above the ground is modeled by h=-16t^2+32t+48. a

Komok [63]

Answer:

a) 64 feet

b) 3 seconds

Step-by-step explanation:

a)

The maximum height of $h=h(t)$ can be bound by finding the y-coordinate of the vertex of $y=-16x^2+32x+48$ .

Compare this equation to $y=ax^2+bx+c$ to find the values of $a,b,\text{ and } c$ .

$a=-16$

$b=32$

$c=48$ .

The x-coordinate of the vertex can be found by evaluating:

$\frac{-b}{2a}=\frac{-32}{2(-16)$

$\frac{-b}{2a}=\frac{-32}{-32}$

$\frac{-b}{2a}=1$

So the x-coordinate of the vertex is 1.

The y-coordinate can be found be evaluating $y=-16x^2+32x+48$ at $x=1$ :

$y=-16(1)^2+32(1)+48$

$y=-16+32+48$

$y=16+48$

$y=64$

So the maximum height of the rocket is 64 ft high.

b)

When the rocket hit's the ground the height that the rocket will be from the ground is 0 ft.

So we are trying to find the second t such that:

$0=-16t^2+32t+48$

I'm going to divide both sides by -16:

$0=t^2-2t-3$

Now we need to find two numbers that multiply to be -3 and add to be -2.

Those numbers are -3 and 1 since (-3)(1)=-3 and (-3)+(1)=-2.

$0=(t-3)(t+1)$

This implies we have either $t-3=0$ or $t+1=0$

The first equation can be solved by adding 3 on both sides: $t=3$ .

The second equation can be solved by subtracting 1 on both sides: $t=-1$ .

So when $t=3$ seconds, is when the rocket has hit the ground.

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