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3 years ago

Start at 5 3/4 subtract 3/8

Mathematics

2 answers:

Eduardwww [97]3 years ago

6 0

Answer:

$53 \div4 - 3 \div 8$

$5 \times 3 = 15 + 4 = 19 \div 4 - 3 \div 8$

$4.75 - 0.375$

$= 4.375$

Lady_Fox [76]3 years ago

4 0

You change the fraction so that they will have the same denominator. This makes it 5 6/8 minus 3/8. The answer is 5 3/8

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Katie has read 32% of a book. if she has read 80 pages, how many more pages does Katie have left to read?

Blababa [14]

Katie has read 32%, so 32%=80 pages. To get the total number of pages, you have to divide 32 and 80 by 8 to get 4% and 10 pages. 4% of the book is 10 pages, and in order to get 100%, you have to multiply by 25(4*25=100) on both sides. Remember, what you do to one side, you do to the other side. That leaves you with 100% and 250 pages. The book is 250 pages long. If she has read 80 pages, you subtract 80 from 250 to find how many pages she has left to read. The difference is 170. Katie has 170 pages left to read.

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3 years ago

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:

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.

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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