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2 years ago

If x is equal to 2 minus root 3 find the value of x minus one upon X whole cube

1 answer:

8 0

"2 minus root 3" => 2 - sqrt(3)

x does not appear here. That could cause a problem.

Unsure of what you mean by "x minus one upon X whole cube." Would you please go back to the original question and ensure that you have copied it down precisely. Without "x" in your equation, there's nothing to solve for.

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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

I need help with my math!!!!!!!

olga_2 [115]

Answer:

2 $\sqrt{6}$

Step-by-step explanation:

Area of a square = $s^{2}$ (s is a side length)

24 = $s^{2}$

$\sqrt{24}$ = $\sqrt{s^{2} }$

$\sqrt{24}$ = s

$\sqrt{(4)(6) }$ = s

2 $\sqrt{6}$ = s

7 0

2 years ago

Please help ASAP+ brainliest

Maurinko [17]

Answer:

Step-by-step explanation:

4 0

2 years ago

A gallery has 25 paintings in its permanent collection, with display space for 10 at one time. How many different collections ca

Anastaziya [24]

You need to know in how many ways you can choose 10 out of 25 items when order does not matter. This is a combination problem.

$_{n}C_{r} = \dfrac{n!}{(n - r)!r!}$

$_{25}C_{10} = \dfrac{25!}{(25 - 10)!10!}$

$_{25}C_{10} = \dfrac{25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15!}{15! \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$_{25}C_{10} = \dfrac{25 \times 24 \times 23 \times 22 \times 21 
\times 20 \times 19 \times 18 \times 17 \times 16}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 
\times 2 \times 1}$

$_{25}C_{10} = \dfrac{23 \times 11 \times 20 \times 19 \times 2 \times 17}{1}$

$_{25}C_{10} = 23 \times 11 \times 20 \times 19 \times 2 \times 17$

$_{25}C_{10} = 3,268,760$

6 0

3 years ago

Read 2 more answers

Plssss help quick i need it for sum important

Vanyuwa [196]

(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

Step-by-step explanation:

(a)

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

7 0

2 years ago