Find the degree of the polynomial: 7.5x^3yz+x^7z^2+2x^3y^3z−x^4z

Please help me on this one

tester [92]

south America = 2.2%

2.2% = 0.022

0.022 = 22/1000 which reduces to 11/500

4 0

3 years ago

For the function F defined by F(x) = x2 – 2x + 4, find F(| — 4|).

Lostsunrise [7]

F(x)=x^2-2x+4

$\\ \sf\longmapsto F(|-4|)$

$\\ \sf\longmapsto F(4)$

$\\ \sf\longmapsto 4^2-2(4)+4$

$\\ \sf\longmapsto 16-8+4$

$\\ \sf\longmapsto 8+4$

$\\ \sf\longmapsto 12$

7 0

2 years ago

The volume of a cube, in cubic centimeters, is given by the function V(x) = x3, where x is the side length of the cube in centim

Sav [38]

I think we can all agree that 1 centimeter=10millimeters.

And if x is the side in centimeters then 10x will be the side in millimeters.

We just need to rewrite x^3 into (10x)^3.

Because you need to do the exponent to all part of the ( )...

(10x)^3=1000x^3

V(x)=1000x^3 cubic millimeters is the answer

5 0

3 years ago

Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin9t,sin6t,9t⟩,r(0)=⟨4,6,3⟩

klemol [59]

$\mathbf r'(t)=\langle\sin9t,\sin6t,9t\rangle$

$\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt$

$\mathbf r(t)=\left\langle\displaystyle\int\sin9t\,\mathrm dt,\int\sin6t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle$

$\displaystyle\int\sin9t\,\mathrm dt=\frac19\cos9t+C_1$

$\displaystyle\int\sin6t\,\mathrm dt=\frac16\cos6t+C_2$

$\displaystyle\int9t\,\mathrm dt=\frac92t^2+C_3$

With the initial condition $\mathbf r(0)=\langle4,6,3\rangle$ , we find

$\dfrac19\cos0+C_1=4\implies C_1=\dfrac{35}9$

$\dfrac16\cos0+C_2=6\implies C_2=\dfrac{35}6$

$\dfrac92\cdot0^2+C_3=3\implies C_3=3$

So the particular solution to the IVP is

$\mathbf r(t)=\left\langle\dfrac19\cos9t+\dfrac{35}9,\dfrac16\cos6t+\dfrac{35}6,\dfrac92t^2+3\right\rangle$

4 0

3 years ago

add the term that makes the given expression into a perfect square. write the result as the square of a bracketed expression c s

Molodets [167]

<h2>Perfect Squares</h2>

Perfect square formula/rules:

$a^2+2ab+b^2=(a+b)^2$
$a^2-2ab+b^2=(a-b)^2$

Trinomials are often organized like $ax^2+bx+c$ .

The <em>b</em> value in this case is <em>c</em>, and it will always equal the square of half of the <em>b</em> value.

Perfect square trinomial: $ax^2+bx+(\dfrac{b}{2})^2$
or $ax^2-bx+(\dfrac{b}{2})^2$

<h2>Solving the Question</h2>

We're given:

$c^2-4c$

In a trinomial, we're given the $ax^2$ and $bx$ values. <em>a</em> in this case is 1 and <em>b</em> in this case is 4. To find the third value by dividing 4 by 2 and squaring the quotient:

4 ÷ 2 = 2
2² = 4

Therefore, the term that we can add is + 4.

$c^2-4c+4$

To write this as the square of a bracketed expression, we can follow the rule $a^2-2ab+b^2=(a-b)^2$ :

$(c-2)^2$

<h2>Answer</h2>

$c^2-4c+4$

$(c-2)^2$

4 0

2 years ago