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Physics-

General

Easy

Question

Which of the following is equal to $open parentheses 1 minus sin to the power of 2 end exponent invisible function application theta close parentheses sec to the power of 2 end exponent invisible function application theta$

1
–1
0
$\frac{1}{2}$

The correct answer is: 1

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Value of $tan invisible function application theta equals fraction numerator 2 over denominator square root of 5 end fraction comma$ find the value of sin²0^O +cos²0^O

Value of $tan invisible function application theta equals fraction numerator 2 over denominator square root of 5 end fraction comma$ find the value of sin²0^O +cos²0^O

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parallel

The density of a material is 8 g/cc. In a unit system in which the unit length is 5 cm and unit mass is 20g, what is the density of the material?

The density of a material is 8 g/cc. In a unit system in which the unit length is 5 cm and unit mass is 20g, what is the density of the material?

physics-General

The density of a material is 8 g/cc. In a unit system in which the unit length is 5 cm and unit mass is 20g, what is the density of the material?

The density of a material is 8 g/cc. In a unit system in which the unit length is 5 cm and unit mass is 20g, what is the density of the material?

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In the given figure, if $P Q perpendicular P S comma P Q parallel to S R comma angle S Q R equals 28 to the power of ring operator end exponent text end text$ and $text end text angle Q R T equals 65 to the power of ring operator end exponent comma$ then the values of x and y respectively are

In the given figure, if $P Q perpendicular P S comma P Q parallel to S R comma angle S Q R equals 28 to the power of ring operator end exponent text end text$ and $text end text angle Q R T equals 65 to the power of ring operator end exponent comma$ then the values of x and y respectively are

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In the given figure, if lines PQ and RS intersect at point T, such that $angle P R T equals 40 to the power of ring operator end exponent comma angle R P T equals 95 to the power of ring operator end exponent text end text$ and $text end text angle T S Q equals 75 to the power of ring operator end exponent comma text then end text angle S Q T$

In the given figure, if lines PQ and RS intersect at point T, such that $angle P R T equals 40 to the power of ring operator end exponent comma angle R P T equals 95 to the power of ring operator end exponent text end text$ and $text end text angle T S Q equals 75 to the power of ring operator end exponent comma text then end text angle S Q T$

In the given figure, if $A B parallel to D E comma angle B A C equals 35 to the power of ring operator end exponent text end text$ and $text end text angle C D E equals 53 to the power of ring operator end exponent$ , then DCE is

In the given figure, if $A B parallel to D E comma angle B A C equals 35 to the power of ring operator end exponent text end text$ and $text end text angle C D E equals 53 to the power of ring operator end exponent$ , then DCE is

In the given $capital delta X Y Z comma angle X equals 62 to the power of 0 end exponent comma angle X Y Z equals 54 to the power of ring operator end exponent$ If YO and ZO are the bisectors of ÐXYZ and ÐXZY respectively, then ÐOZY and ÐYOZ are

In the given $capital delta X Y Z comma angle X equals 62 to the power of 0 end exponent comma angle X Y Z equals 54 to the power of ring operator end exponent$ If YO and ZO are the bisectors of ÐXYZ and ÐXZY respectively, then ÐOZY and ÐYOZ are

parallel

In the given figure, sides QP and RQ of $capital delta P Q R$ are produced to points S and T respectively If $angle S P R equals 135 to the power of ring operator end exponent text and end text angle P Q T equals 110 to the power of ring operator end exponent$ then $angle text PRQ end text$ is

In the given figure, sides QP and RQ of $capital delta P Q R$ are produced to points S and T respectively If $angle S P R equals 135 to the power of ring operator end exponent text and end text angle P Q T equals 110 to the power of ring operator end exponent$ then $angle text PRQ end text$ is

In the given figure, an equilateral triangle EDC surmounts the square ABCD then the value of x is

In the given figure, an equilateral triangle EDC surmounts the square ABCD then the value of x is

In the given figure, AD BC $perpendicular$ and AE is the bisector of $angle$ BAC Then $angle$ DAE is

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In the given figure, XY is parallel to PQ, then the measures of the angles x⁰ and y⁰ are

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In the given figure the sides AB and AC of a triangle are produced The bisectors of exterior angles B and C so formed intersect each other at point O $angle$ BOC is

In the given figure the sides AB and AC of a triangle are produced The bisectors of exterior angles B and C so formed intersect each other at point O $angle$ BOC is

In the given figure, the bisectors of $angle$ Q and $angle$ R meet at the point O then $angle$ QOR is

In the given figure, the bisectors of $angle$ Q and $angle$ R meet at the point O then $angle$ QOR is

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