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zhuklara [117]
3 years ago
11

Is psat 8/9 easy please tell me i have to take it

Advanced Placement (AP)
2 answers:
Zanzabum3 years ago
8 0

Answer:

It depends if you know what to do

Explanation:

Hope this helps

Elodia [21]3 years ago
3 0

I hate to be this type of person, but it really depends on whether you understand the stuff you have been taught. As well as whether you have studied. It can be easy if you put effort into making sure you know the materials for the quiz, but also can be hard if you jump into it blindly.

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Pag tama ng panganib​
Zinaida [17]
Wait is this Spanish ? Or you just giving free points ?
4 0
3 years ago
A biker goes from 15 ms -1to 0 ms -1after hitting a brick wall. Its rate of acceleration is -50ms -2 . Calculate the time taken
Damm [24]

Answer:

Time taken t = 4.167 seconds (Approx)

Explanation:

Given:

Initial Speed u = 12 m/s

Final speed v = 0 m/s

Acceleration a = -50 m/s²

Find:

Time taken t

Computation:

v = u + at

0 = 12 + (-50)t

t = 4.167 s

Time taken t = 4.167 seconds (Approx)

4 0
3 years ago
PLEASE HELPPP
otez555 [7]

Answer:

D.Green building......

3 0
3 years ago
(25 POINTS) PLEASE HELP WILL GIVE BRAINLIEST, THANKS AND 5 STAR RATING!!!
bulgar [2K]

PYTHAGOREAN THEOREM\\c^{2} = a^{2} + b^{2}\\ b^{2}=c^{2} -a^{2}\\ a^{2} =b^{2}-c^{2}

1) x=9, because we can use the Pythagorean theorem, x^2 + 144 = 225, x^2 = 81

2) x=3\sqrt{8}, because 6^2 +6^2 = 72, \sqrt{72} = 3\sqrt{8}

3) x=2\sqrt{3}, because x^{2} +7=19\\x^{2} =12\\x= 2\sqrt{3}

hope this helps... load for images!

5 0
3 years ago
I can’t figure this problem out
Natalka [10]

Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

\int\limits^4_1 {(3f(x)+2)} \, dx

So lets break this up into two different integrals that represent the same area.

\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

\int\limits^4_1 {f(x)} \, dx

But since we have these values we can say that:

-3 - 2 = -5

Which means that \int\limits^4_1 {f(x)} \, dx = -5

So now we can evaluate \int\limits^4_1 {(3f(x)+2)} \, dx

Lets first break up our integrand into two integrals

\int\limits^4_1 {(3f(x)+2)} \, dx = 3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx

Now we can evaluate this:

We know that \int\limits^4_1 {f(x)} \, dx = -5

So:

3(-5)+2[x] where x is evaluated at 4 to 1 so

-15 + 2(3)

So we are left with -15 + 6 = -9

5 0
3 years ago
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