All categories

3 years ago

Which statement is true about the right angles ABC and DEF

Mathematics

1 answer:

cluponka [151]3 years ago

7 0

The answer is the first one

You might be interested in

Factor the expression by finding the GCF. 16m2 − 12m

ziro4ka [17]

Answer:

4m(4m-3)

Step-by-step explanation:

16m^2 − 12m

16 m^2 = 2*2*2*2*m*m

12m = 2*2*3*m

The greatest factor for 16m^2 and 12m is 2*2*m or 4m

Factor out 4m

16 m^2 = 2*2*2*2*m*m = 4m(2*2m)=4m(4m)

12m = 2*2*3*m = 4m(3)

Factoring out 4m

16m^2 − 12m

4m(4m-3)

3 0

3 years ago

Read 2 more answers

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.

The inner surface: $Si=4\pi (0.06m)^{2} = 0.04 m^{2}$

The outer surface: $S=4\pi (0.08m)^{2}=0.08m^{2}$

Now we can calculate the charges,

Inner charge: $Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC$

Outer charge: $Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC$

Then, we are able to calculate both fields:

Inner field: $Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}$

Outer field: $Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}$

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0

3 years ago

How do you solve his with working

AlexFokin [52]

Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

$\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888$

b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

$\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884$

7 0

3 years ago

Line L passes through the points (0, --3) and (6,9).<br> (a) Find the equation of line L.

valentina_108 [34]

Step-by-step explanation:

Given points are :

$(0,\:-3)=(x_1,\:y_1) \:\&\: (6,\:9)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -( - 3)}{ -3 -9 } = \frac{x -0 }{0 -6 } \\ \\ \therefore \frac{y + 3}{ -12} = \frac{x }{ -6 } \\ \\ \therefore \frac{y + 3}{ 2} = \frac{x }{1} \\ \\ \therefore \: y + 3 = 2x \\ \\ \huge \red{ \boxed{\therefore \: 2x - y - 3 = 0}} \\ is \: the \: required \: equation \: of \: line.$

5 0

3 years ago

I need help on this help help I’ll give out some money

frozen [14]

<h3>a) Never</h3>

{All angles of a rectangle are right}

<h3>b) Always</h3>

{all sides of a rhombus are the same, 4×13=52}

<h3>c) Always</h3>

{oposite angles of a paralleogram are congruent}

<h3>d) Never</h3>

{parallel sides has the same slope}

<h3>e) Always</h3>

{square has all sides of the same length, so it is rhombus}

<h3>f) Sometimes</h3>

{Only if it has angles of 90°}

6 0

2 years ago