Simplify 6.4a - 4.7a

Mathematics

2 answers:

sveta [45]3 years ago

4 0

4.7a+8.2a-6.5a=0

We simplify the equation to the form, which is simple to understand

4.7a+8.2a-6.5a=0

We move all terms containing a to the left and all other terms to the right.

+4.7a+8.2a-6.5a=0

We simplify left and right side of the equation.

+6.4a=0

We divide both sides of the equation by 6.4 to get a.

a=0

docker41 [41]3 years ago

3 0

$6.4a-4.7a\\6.4a+-4.7a\\=1.7a$

Answer is 1.7a

Hope it helped.

-Charlie

You might be interested in

(PLEASE ANSWER AM TIMED! WILL MARK RIGHT ANSWER AS BRAINLIEST AND WILL THANK AND RATE!)

olganol [36]

Answer:

Well, Your best answer will be is 225° rotation. Because, The points A' and C' are tilted more than 180 degrees from the origin of rotation. If it were 90 degrees, the A' and C' would have been in the right side. So, The best answer will be is 225°. 

Hope That Helps!

$^{Itsbrazts}$

5 0

3 years ago

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$