Solve the quadratic equation below using the Quadratic Formula $x squared plus 5 x minus 14 equals 0$

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.

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 $text end text fraction numerator B D over denominator A B end fraction equals$

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.

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

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

In the figure, AB = BC = CD = DE = EF and AF = AE. Then $angle A$ = ?

In the figure, AB = AC, $angle A$ = 48° and $angle A C D$ = 18° . Then

an isosceles triangle is one that has two of its sides equal and the base angles equal to each other.

In the figure, AB = AC, $angle A$ = 48° and $angle A C D$ = 18° . Then

an isosceles triangle is one that has two of its sides equal and the base angles equal to each other.

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of $triangle A B C$ . Then x° + y° is

if one of the angles were acute then we'd not have been able to conclude whether the sum of the angles were greater than or less than 180 degrees.

In the figure, ‘x° ’ and ‘y° ’ are two exterior angle measures of $triangle A B C$ . Then x° + y° is

if one of the angles were acute then we'd not have been able to conclude whether the sum of the angles were greater than or less than 180 degrees.

The perimeter of the following figure is

perimeter is the sum of the sides of a polygon.

The perimeter of the following figure is

perimeter is the sum of the sides of a polygon.

chemistry-

Two pupils Ram and Shyam have taken lime water in a vessel, as shown in the figure. Ram is inhaling air from straw A and Shyam is blowing air through straw B. In which case lime water turns milky faster and Why?

chemistry-General

If $straight A equals open square brackets a subscript i j end subscript close square brackets subscript 2 cross times 2 end subscript$ where $a subscript i j end subscript equals i plus j$ then A is equal to