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2 years ago

What is the area of the figure

Mathematics

1 answer:

Tomtit [17]2 years ago

4 0

A²+b²=c²
a=b because it has 2 45° angles and one 90 so it is isosceles right triangle

∴ a² + a² = c²
2a²=24²
2a²=576
a²=288
a=√288
a=16.97056275

Therefore. With A=1/2bh
A=(1/2) * 16.97056275 * 16.97056275
= (1/2) * 288
= 144

144 ft²

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PLEASEEE HELP ! I WOULD APPRECIATE IT VERY MUCH :)

azamat

hi!

so the original equation is (x^2 -121) / x+ 11.

Those two equations look similar...

well, we know that the top equation looks like a^2 - b^2, and that equation iis equivalent to this equation: (a+b)(a-b).

So if we factor that out, we get:

( (x + 11) * (x-11) )/ (x + 11)

we can cancel the x+11 on the top and the x+11 on the bottom out.

that leaves us with x-11.

Hope this helped!

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2 years ago

4. What is the value?<br> 5.function or not?

Vaselesa [24]

1. No.
2. No.
3. Yes.
4. No.

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3 years ago

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi

ANTONII [103]

Answer:

It takes 3 seconds over the interval [0,3]

Step-by-step explanation:

To find when the roller coaster reaches the ground, find when d=0.

$0=144-16t^2$

To solve divide each term by 16 and factor:

$\frac{0}{16}=\frac{144}{16} -\frac{-16t^2}{16} \\0= 9 - t^2\\0=(3-t)(3+t)$

Solve for t by setting each factor to 0.

t-3=0 so t=3

t+3=0 so t=-3

This means the car is in the air from 0 to 3 second.

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3 years ago

I need help on this question

raketka [301]

You got the right answers!
Horizontal and vertical

Congrats

7 0

3 years ago

a group of astronauts launched a model rocket from a platform. Its flight path is modeled by h= -4t^2+24t+13 where h is the heig

Eddi Din [679]

Answer:

6.5 seconds

Step-by-step explanation:

Keep in mind that when $h=0$ , this is the same height for both when the model rocket takes off and lands, so when the rocket lands, time is positive. Thus:

$h=-4t^2+24t+13\\\\0=-4t^2+24t+13\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-24\pm\sqrt{24^2-4(-4)(13)}}{2(-4)}\\ \\t=\frac{-24\pm\sqrt{576+208}}{-8}\\\\t=\frac{-24\pm\sqrt{576+208}}{-8}\\\\t=\frac{-24\pm\sqrt{784}}{-8}\\\\t=\frac{-24\pm28}{-8}\\\\t=\frac{-24-28}{-8}\\ \\t=\frac{-52}{-8}\\ \\t=\frac{52}{8}\\\\t=6.5$

So, the amount of seconds that the model rocket stayed above the ground since it left the platform is 6.5 seconds

5 0

2 years ago

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