All categories

3 years ago

Write the following as a complex number in the form a + bi square root -245

Mathematics

2 answers:

Pepsi [2]3 years ago

5 0

Answer:

Step-by-step explanation:

√(-245)=√(49×5×i^2)=7√5 i=0+7√5 i

svetoff [14.1K]3 years ago

3 0

Answer:

$\sqrt{-245}$ in the form a + bi is $0+7\sqrt{5}i$

Step-by-step explanation:

We know that

$i=\sqrt{-1}$

We need to find $\sqrt{-245}$

$\sqrt{-245}=\sqrt{-1}\times \sqrt{245}=i\sqrt{245}=i\sqrt{49\times 5}=i\times 7\sqrt{5}$

$\sqrt{-245}$ in the form a + bi is $0+7\sqrt{5}i$

You might be interested in

Can someone please help with finding the surface area..

Vsevolod [243]

Answer:

it is 20.5

Step-by-step explanation:

you need to add all of the numbers together

5 0

2 years ago

Read 2 more answers

Find all points on the curve y = cos x, 0 ≤ x ≤ 2π, where the tangent line is parallel to the line y = -x.

andre [41]

Answer:

90⁰

Step-by-step explanation:

y=cosx

y'= -sinx

eqñ of tangent is y=-x

slope of tangent =-1

we know...

slope of tangent = y' or derivative of y

-sin x = -1

sin x= 1

x = 90⁰

7 0

2 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

3 years ago

The local bike shop sells a bike and accessories package for $266. If the bike is worth 6 times more than the accessories, how m

tatiyna

Do 6x 266 in your calculator that’s ur answer

7 0

3 years ago

Find the surface area of the composite solid and round to the nearest tenth. Explain your answer.

Orlov [11]

Well, notice the composite is really just 4 triangles atop sitting on top of 4 rectangles, and all of them area stacked up at the edges.

so, for the rectangle's sides,

front and back are two 6x3 rectangles

left and right are two 6x3 rectangles

the bottom part is a 6x6 rectangle

now, we don't include the 6x6 rectangle that's touching the triangles, because that's inside area, and is not SURFACE area, so we nevermind that one.

now, the triangles are just four triangles with a base of 6, and a height of 4, in red noted there.

so, just get the area of all those rectangles and the triangles, sum them up and that's the surface area of the composite,

$\bf \stackrel{lateral~rectangles}{4(6\cdot 3)}~~+~~\stackrel{bottom~rectangle}{(6\cdot 6)}~~+~~\stackrel{triangles}{4\left( \cfrac{1}{2}(6)(4) \right)}$

7 0

2 years ago