•Which is the BEST indication that a chemical reaction has taken place?A.Blue powder turns water blue.B. A white substance dissolving in water. C. Metal bubbling when placed in an unknown liquid. •What are the products in the reaction: CH + 20,CO, +2H,0 A. Carbon Dioxide and Water. B. Carbon Monoxide and Hydroxide.C. Salt and Vinegar. •Which of the following is an example of a chemical reaction? A. salt crystals forming as sea water evaporates.B. paper turning soft when wet. C. leaves changing colours in the fall.tell me this MCQS answer very argent.​

Answer:

[tex]इनमे \: से \: कोई \: नही[/tex]

Explanation:

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स्पष्टीकरण ⦂

वाहिकाएं तथा वाहिनिकायें जाइलम की तत्व होती हैं। जाइलम एक प्रकार का जटिल ऊतक होता है। जो अनेक प्रकार की कोशिकाओं से बना होता है। जाइलम के मुख्य भागों में जाइलम तंतु, जाइलम पेरेन्काइमा, वाहिकाएं और वाहिनिकायें होती हैं। ये चार भाग जाइलम के मुख्य भाग होते हैं। जाइलम का मुख्य कार्य पौधे में जड़ से पत्तियों तक जल और खनिज लवण का संवहन करना है।

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Step-by-step explanation:

only 2 tangents can be drawn to a circle

hope it helps u

Answer:

Line joining the center of the circle to the center of the chord would be a perpendicular bisector to the chord. Another way of solving this question is directly by constructing a circle with the given radius and then draw an arc with the given chord length then measure the distance.

Step-by-step explanation:

1. Construct a circle of radius AB = 3.5 cm.
2. Make a point P at a distance of OP = 6.2 cm, and join OP.
3. Draw a perpendicular bisector of OP.  
4. Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at X and Y.
5. Join PX and PY to obtain the required tangent
6. Thus, PX and PY are the required tangents.

Let us name the center of the chord as shown in the above diagram as [tex]\mathrm{E}$.[/tex]

So now, [tex]$\angle A E D=\angle A E B=90^{\circ}$[/tex]As the line [tex]$\mathrm{AE}$[/tex] would be the perpendicular bisector of the chord [tex]$\mathrm{BD}$[/tex].

We know that the radius of the circle is [tex]3.5 \mathrm{~cm}$.[/tex] let us assume [tex]r=3.5 \mathrm{~cm}$.[/tex]

[tex]$\mathrm{AD}=\mathrm{AB}=3.5 \mathrm{~cm}$[/tex]

[tex]$\mathrm{AE}$[/tex] is the bisector of the line segment [tex]$\mathrm{BD}$[/tex]

[tex]$$B E=E D=\frac{B D}{2}$$[/tex]

As given in the question the length of chord [tex]$\mathrm{BD}$[/tex] is [tex]$\mathrm{BD}=6 \mathrm{~cm}$[/tex]

[tex]$B E=E D=\frac{B D}{2}=\frac{6}{2}=3 \mathrm{~cm}$$[/tex]

[tex]\mathrm{ED}=3 \mathrm{~cm}$\\$\angle A E D=90^{\circ}, \triangle A D E$[/tex] would be a right angled triangle with two of its sides as[tex]$\mathrm{ED}=3 \mathrm{~cm}, \mathrm{AD}=3.5 \mathrm{~cm}$[/tex]

Pythagoras theorem can be used as [tex]$\angle A E D=90^{\circ}$[/tex]

From Pythagoras theorem,[tex]$(A D)^{2}=(E D)^{2}+(E A)^{2}$[/tex]

Substituting the values of [tex]$\mathrm{ED}=3 \mathrm{~cm}$[/tex] and [tex]$\mathrm{AD}=3.5[/tex][tex]\mathrm{~cm}$[/tex] we can obtain the value of [tex]$\mathrm{EA}$[/tex]

[tex]$$\begin{aligned}&(A D)^{2}=(E D)^{2}+(E A)^{2} \\&\Rightarrow(3.5)^{2}=(3)^{2}+(E A)^{2} \\&\Rightarrow(E A)^{2}=(3.5)^{2}-(3)^{2} \\&\Rightarrow(E A)^{2}=12.25-9 \\&\Rightarrow(E A)^{2}=3.25 \\&\Rightarrow(E A)=\sqrt{3.25} \\&\Rightarrow(E A)=1.8027 \\&\mathrm{EA}=1.8027 \mathrm{~cm}\end{aligned}$$[/tex]

Therefore EA is the distance between the centre of the chord and centre of the circle, which is EA = [tex]$1.8027 \mathrm{~cm}$[/tex]

Answer:

1/3

Explanation:

I just gave it a try

if it is correct,

then tell me in the comments

hope it help you

thank you

For the given network, the effective resistance between A and B is R₁ and between A and D is R₂.

We have to find the value of R₁/R₂.

Case 1 : effective resistance between A and B = R₁

Here we see, two resistors of same resistance R are joined in series.

∴ r' = R + R = 2R

Now, two resistors of resistance R, four resistors of resistance 2R are joined in series.

∴ r'' = R + R + 2R + 2R + 2R + 2R = 10R

Now, r' and r'' are joined in parallel.

∴ effective resistance, R₁ = r'r''/(r' + r'') = (2R × 10R)/(2R + 10R) = (5/3)R

Case 2 : effective resistance between A and D = R₂

A to C, there are four resistors of same resistance R and from C to D, one resistor of resistance 2R are joined in series.

∴ R' = R + R + R + R + 2R = 6R

from A - F - E - D , there are three resistors of same resistance 2R are joined in series.

∴ R'' = 2R + 2R + 2R = 6R

Now, R' and R'' are joined in parallel.

∴ effective resistance, R₂ = R'R''/(R' + R'') = (6R × 6R)/(6R + 6R) = 3R

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a) R1 = R2

b) R1 > R2

c) R1 < R2

d) R1 = R2 = 0

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Answer:

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