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2 years ago

Help plz asap i reward brainliest im being timed!!!!

Mathematics

2 answers:

Bumek [7]2 years ago

8 0

I looked it up on desmos, I am positive that this is the correct answer

Lera25 [3.4K]2 years ago

3 0

It should be 4 I think

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Help me please<br> 8. Match each angle pair with the corresponding drawing. Choose the best answer!

sdas [7]

Answer:

1. D

2. B

3. A

4. C

Step-by-step explanation:

1. Vertical angles are a pair of opposite angles formed by the intersection of lines. Choice D has two lines intersecting, and two opposite angles formed by it.

2. Adjacent angles are two angles that have a common vertex and common side but do not overlap. Choice B is two angles with a common vertex and common side, and they don't overlap.

3. Supplementary angles are angles that add up to 180 degrees/make a straight angle. Choice A has two angles that add up to 180 degrees, and the two angles combined make a straight angle/line.

4. Two angles are called complementary angles if they add up to 90 degrees and form a right angle. Choice C has two angles that form a right triangle, meaning it adds up to 90 degrees.

5 0

2 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

2 years ago

Evaluate. −27/125−−−−√3 A.-9/25 B.-3/5 C.9/25 D.3/5

Vladimir79 [104]

$\sqrt[3]{\dfrac{-27}{125}} = -\sqrt[3]{\left(\dfrac{3}{5}\right)^{3}}=\dfrac{-3}{5}$

The appropriate choice appears to be ...
B. -3/5

7 0

3 years ago

Read 2 more answers

(02.01 LC)

ExtremeBDS [4]

Answer:

Yes it is a function

Step-by-step explanation:

We have to check the ordered pairs to find out if given relation is a function or not.

In an ordered pair, the first element represents the input and the second element represents the output.

The set of inputs is domain and output is range.

For a relation to be function, there should be no repetition in domain i.e there should be unique pairs of input and output.

In the given relation, the domain is {3,5,-1,-2}.

No element is repeated hence it is a function ..

7 0

2 years ago

1. Using the above

Ivan

Answer:

yes there is a correlation

Step-by-step explanation:

here's how I found this by looking up all key words.

4 0

1 year ago