The magnetic field at the centre of circular loop in the circuit shown below is

physics-

A thin 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.45T magnetic field as shown in Fig .A battery and a 25 $capital omega blank$ resistor in series are connceted to the supports. The largest voltage the battery can have without breaking the circuit at the supports (units are in ”V”) is

physics-General

Solve the quadratic equation below using the Quadratic Formula $x^{2} + 5 x - 14 = 0$

GeneralGeneral

In the figure, $C A B equals 90 to the power of ring operator end exponent text and end text A D perpendicular B C.$ If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD =

we can use the property that ratio of sides remains same in similar triangles.

In the figure, $C A B equals 90 to the power of ring operator end exponent text and end text A D perpendicular B C.$ If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD =

we can use the property that ratio of sides remains same in similar triangles.

$text In the figure end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text and end text A B equals 5 c m. text The value of end text D C text is end text$

we can use the property that ratio of sides remains same in similar triangles.

$text In the figure end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text and end text A B equals 5 c m. text The value of end text D C text is end text$

we can use the property that ratio of sides remains same in similar triangles.

maths-

In the figure $text , end text stack D E with bar on top divided by divided by stack B C with bar on top text end text$ and area $text end text left parenthesis triangle A D E right parenthesis equals text area end text left parenthesis triangle B C E D right parenthesis. text end text$ The value of $text end text fraction numerator B D over denominator A B end fraction equals$

maths-General

In the figure, $angle A equals angle C E D$ , AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

In the figure, $angle A equals angle C E D$ , AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

In the figure $text , end text stack A B with bar on top divided by divided by stack Q R with bar on top text . If end text A B equals 3 c m comma P B equals 2 c m text and end text$ PR = 6 cm then QR = ? cm

we can use the property that ratio of sides remains same in similar triangles

In the figure $text , end text stack A B with bar on top divided by divided by stack Q R with bar on top text . If end text A B equals 3 c m comma P B equals 2 c m text and end text$ PR = 6 cm then QR = ? cm

we can use the property that ratio of sides remains same in similar triangles

In the figure $text , end text stack A B with bar on top divided by divided by stack C D with bar on top text and end text stack A C with bar on top intersection stack B D with bar on top equals 0. text If end text O A equals 3 x minus 1 text , end text$ OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.

In the figure $text , end text stack A B with bar on top divided by divided by stack C D with bar on top text and end text stack A C with bar on top intersection stack B D with bar on top equals 0. text If end text O A equals 3 x minus 1 text , end text$ OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

Two line segments $text end text stack A B with rightwards arrow on top text and end text stack C D with bar on top text end text$ intersect at E such that $triangle A C E tilde operator triangle B D E.$ If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

Two line segments $text end text stack A B with rightwards arrow on top text and end text stack C D with bar on top text end text$ intersect at E such that $triangle A C E tilde operator triangle B D E.$ If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

maths-

From the adjacent figure ,the values of x and y are

maths-General

In the given figure, ABC is an Isosceles Triangle in which AB = AC, then

we can also use trigonometry to solve this problem. For the same perpendicular distance from the vertex, the length of side increases with increase in the vertex angle since the perpendicular is the cosine component of the side.
height = side x cos(theta). we know that cosine is a decreasing function. therefore, to keep the height same, the side has to increase.

In the given figure, ABC is an Isosceles Triangle in which AB = AC, then