Hey guys please answer me this ASAP

Vanyuwa [196]

Gabriel and Jorge have a right triangle

Makayla has a equilateral

Emma has an obtuse

4 0

2 years ago

The length of the hypotenuse of a right triangle is 145 units. The length of one leg of the triangle is 144 units. Mike wrote th

ladessa [460]

Answer:

A

Step-by-step explanation:

a^2 + 144^2 = 145^2

a^2 + 20736 = 21025

a^2 = 289

a = sqrt(289)

7 0

2 years ago

Karen make $5 per hour babysitting and $12 per hour giving music lessons. On the weekend, she worked a total of 18 hours and mad

Lilit [14]

Answer:

11 hours

Step-by-step explanation:

Let's say Karen makes x hours babysitting and y hours giving music lessons.

Because she worked a total of 18 hours, we know that x + y = 18.

5 * x is how much money she earns babysitting for x hours, while 12 * y is how much money she earns giving music lessons for y hours. Adding them up gives her total revenue:

5x + 12y = 139

We now have a system of equations. Let's multiply both sides of x + y = 18 by 12:

12 * (x + y) = 18 * 12

12x + 12y = 216

Now, subtract 5x + 12y = 139 from 12x + 12y = 216:

12x + 12y = 216

- 5x + 12y = 139

_________________

7x + 0y = 77

Divide both sides by 7:

x = 11

Thus, Karen spent 11 hours babysitting.

4 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

Need Help!!!! Please!!! Explain!!!! Don't Understand!!!

Arturiano [62]

90 it is a right tranigle

6 0

2 years ago