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2 years ago

What’s the answer to this maths question

Mathematics

1 answer:

Dima020 [189]2 years ago

5 0

108

find the area of the triangles first:
formula for the area of a triangle is A=bh/2
the base is 12
the height can be found by taking the 13(given) subtracting 5(given) and dividing by 2 which gives you 4 and then just fill in the formula
A= 12•4/2
A=24 then multiply by 2(bc there are two triangles and you get
A=48

now to the area of the rectangle which is just
A=BH you fill it in
A=12•5 solve
A=60
then you add the area of the triangles to the area of the rectangle to get the area of the hexagon
A=48+60
A=108

brainliest pls<3

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Answer:

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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 $a,b,\text{ and } c$ .

$a=-16$

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$c=48$ .

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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