All categories

2 years ago

Which of these can not form a perpendicular bisector?v PLZ HALLLLPPPP

Mathematics

2 answers:

polet [3.4K]2 years ago

4 0

A circle obviously. A perpendicular bisector requires right angles which a circle can physically never have.

jasenka [17]2 years ago

3 0

Answer: Second option is correct.

Step-by-step explanation:

Since we know that to form perpendicular bisector .

We need atleast two lines which form right angle.

So, right triangle can form perpendicular bisector with base and height.

Parallel lines can also form perpendicular bisector if a line is drawn from mid point of one line to mid point of another line.

A line segment can form perpendicular bisector.

But circle can't as it is curved shape it needs line to make perpendicular bisector.

Hence, Second option is correct.

You might be interested in

4x^2+4y^2+20x-16y+37=0 standard form

zhenek [66]

The answer in standard form is (x+5/2)^2+(y-2)^2=1

5 0

2 years ago

Need answer urgently please will give thanks and mark Brainliest

babymother [125]

Answer:

Step-by-step explanation:

sqrt(7) = about 2.6

and that divided by 2 is

0.8 which means D is the Answer

8 0

2 years ago

What is the sum or whatever of this equation. This equation is V = πx3^2x8

navik [9.2K]

Answer:

$V=226.08$

Step-by-step explanation:

To evaluate $V$ from the following expression:

$V=\pi\times3^2\times 8$

Taking $\pi=3.14$

$V=3.14\times 3^2\times 8$

$V=3.14\times 9\times 8$

∴ $V=226.08$

6 0

2 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

2 years ago

Work out the height of a triangle with base b= 7.7mm and area a=164.78mm2

solong [7]

Answer:

21.4 mm

Step-by-step explanation:

area ÷ base = height

164.78 mm^2 ÷ 7.7 mm

21.4 mm

Check:

base • height = area

7.7 mm • 21.4 mm = 164.78 mm^2

5 0

2 years ago