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3 years ago

Order these from greatest to least: 6/5 4/2 2/5 1/2

Mathematics

2 answers:

user100 [1]3 years ago

5 0

Answer: 4/2,1/2,6/5,2/5

Step-by-step explanation:

Think of it like pices of pie if you eat half the pie it going to be more than eating 2 pieces right?

Artyom0805 [142]3 years ago

5 0

Answer:

1. 4/2, 2. 6/5, 3. 1/2, 4. 2/5

Step-by-step explanation:

6/5= 1 1/5 or 1.5
4/2= 2
2/5= .4
1/2= .5

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Sin 40 = x/12 what is the value of x?

eimsori [14]

7,713451316238472
X= 7.71345

4 0

3 years ago

What is x^2 -11x+19=-5 as standard form?

IrinaVladis [17]

Answer:

Standard Form; x^2 - 11x + 24 = 0

Step-by-step explanation:

Simplify rewrite such that the quadratic equation has 0 isolated on the right or left side, most commonly right side;

x^2 - 11x + 19 = - 5 ⇒ Add 5 on either side of equation,

Standard Form; x^2 - 11x + 24 = 0

4 0

3 years ago

Consider the force field and circle defined below.

grin007 [14]

By Green's theorem,

$\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy$

where $C$ is the circle $x^2+y^2=9$ and $D$ is the interior of $C$ , or the disk $x^2+y^2\le1$ .

Convert to polar coordinates, taking

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

Then the work done by $\vec F$ on the particle is

$\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0$

4 0

3 years ago

I need help im gonna post 3 more different questions pls answer them.

defon

Answer:

122

Step-by-step explanation:

1^6 = 1

11^2 = 121

1 + 121 = 122

hope this helps

7 0

3 years ago

Read 2 more answers

Which expression is equivalent to b^-2/ab^-5? a) a/b^5 b)1/ab^5 c)a^3b/1 d)b/a

Snezhnost [94]

$\bf a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
\cfrac{1}{a^{-n}}\implies a^{{ n}}\\\\
-------------------------------\\\\
\cfrac{b^{-2}}{ab^{-5}}\implies \cfrac{b^{-2}\cdot b^{+5}}{a}\implies \cfrac{b^{-2+5}}{a}\implies \cfrac{b^3}{a}$

4 0

3 years ago