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2 years ago

In the figure below, if BE is the perpendicular bisector of AD , what is the value of x?

Mathematics

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

Step-by-step explanation:

<h3>Since BE is perpendicular bisector AE=ED</h3><h3>Substituting we have:</h3><h3>3x + 5 = x + 25 solving for x</h3><h3> x = 10</h3><h3 />

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Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.

Rama09 [41]

Check the picture below.

now, we have a triangle with all three sides, thus we can use Heron's Area Formula on the triangle.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=10\\ b=26.695\\ c=22\\ s=29.3475 \end{cases} \\\\\\ A=\sqrt{29.3475(29.3475-10)(29.3475-26.695)(29.3475-22)} \\\\\\ A=\sqrt{29.3475(19.3475)(2.6525)(7.3475)}\implies A\approx \sqrt{11066.007} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill A\approx 105.195~\hfill$

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2 years ago

A cylinder and a cone have congruent heights and radii.what is the ratio of the volume of the cone to the volume of the cylinder

NemiM [27]

Answer:

1: 3 ratio of the volume of the cone to the volume of the cylinder

Step-by-step explanation:

Volume of cone(V) is given by:

$V = \frac{1}{3} \pi r^2h$

where, r is the radius and h is the height of the cone.

Volume of cylinder(V') is given by:

$V' = \pi r'^2h'$

where, r' is the radius and h' is the height of the cylinder.

As per the statement:

A cylinder and a cone have congruent heights and radii.

⇒r = r' and h = h'

then;

$\frac{V}{V'} = \frac{ \frac{1}{3} \pi r^2h}{\pi r'^2h'} = \frac{ \frac{1}{3} \pi r'^2h'}{\pi r'^2h'}$

⇒ $\frac{V}{V'} =\frac{1}{3} = 1 : 3$

Therefore, the ratio of the volume of the cone to the volume of the cylinder is, 1 : 3

4 0

3 years ago

Read 2 more answers

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Alisiya [41]

$\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
sin(x)-1=0\\
sin(x)=1\to x=sin^{-1}(1)
\end{cases}$

$\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}$

3 0

3 years ago

Starting from 50 feet away your friend skates towards you and then passes you. She skates at a constant speed of 10 feet per sec

Kitty [74]

12 sec..............

6 0

3 years ago

Henry and Blake are both runners. In 2 hours, Henry runs 14 miles. In 5 hours, Henry runs 35 miles. The graph below represents t

Tasya [4]

Henry runs at a rate of seven miles an hour. I think you meant the second sentence to be 'Blake', because there is not attatched graph. They are both running at the same speed in this case.

I got this because 14 (miles ran by Henry) / 2 (hours ran) you get 7. This same equation is applied to (Blake?) 35 (miles ran by [Blake?]) / 5 (hours ran) also equals seven.

If you meant the second statement to be Blake, they are both running at the same speed, and x=7.

If you didn't, then there is not enough information to determine Blake's speed.

If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one, as well as to move up in rank! :)

3 0

3 years ago