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2 years ago

What property is shown in the following equation? 17 + 8 + 3 = 17 + 3 + 8

Mathematics

2 answers:

11111nata11111 [884]2 years ago

7 0

Answer:

Associative property

ss7ja [257]2 years ago

3 0

Answer:

The Commutative Law of Addition.

Step-by-step explanation:

Basically, this law says we can swap numbers over and still get the same number when we add. This has demonstrated exactly that.

Hope this helped!

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I have two proof questions and I don't really know how to do them.

goldfiish [28.3K]

$\angle1\cong\angle2$ corresponding angles, therefore

m∠1 = m∠2 → m∠2 = 130°

$\angle2\cong\angle3$ vertical angles, therefore

m∠2 = m∠3 → m∠3 = 130°

If ∠1 and ∠2 are complementary, then m∠1 + m∠2 = 90°.

If ∠2 and ∠3 are complementary, then m∠2 + m∠3 = 90°

Therfore

$m\angle1 + m\angle2 = m\angle2 + m\angle3\qquad|\text{subtract}\ m\angle2\ \text{ from both sides}\\\\m\angle1 = m\angle3\to\angle1\cong\angle3$

5 0

2 years ago

PLEASE HELP ME SOLVE!

Crank

Answer:

F(-6) = 2*(-6) + 1

5*(-6) -2

= -11

-32

= 11

5 0

2 years ago

Read 2 more answers

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

The value 0.1 meter is equivalent to 1 ____.

r-ruslan [8.4K]

Answer:

centimeters I think so

4 0

3 years ago

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I'm confused about this, please help!

Fofino [41]