Which of the following are correct ?

Number of cyphers after decimal before a significant figure comes in $open parentheses 5 over 3 close parentheses to the power of negative 100 end exponent$ is

In this question, we have to find the number of cyphers. We need to remember that the cypher number can be found from the logarithmic calculation but to find the actual value or to visualize it we need to remove the logarithm. Here, log₁₀5 = 0.6990 and log₁₀3 = 0.4770.

Number of cyphers after decimal before a significant figure comes in $open parentheses 5 over 3 close parentheses to the power of negative 100 end exponent$ is

The value of p for which both the roots of the quadratic equation, $4 x squared minus 20 p x plus open parentheses 25 p squared plus 15 p minus 66 close parentheses$ are less than 2 lies in :

In this question, we have to find where the p lies. Here, we use discriminant, which is $b squared – 4 a c$ . if D > 0 and D = 0 then real solution but if D < 0 then imaginary solution. Here we also us Factorization of quadratic equations.

The value of p for which both the roots of the quadratic equation, $4 x squared minus 20 p x plus open parentheses 25 p squared plus 15 p minus 66 close parentheses$ are less than 2 lies in :

If $log subscript c space 2 times log subscript b space 625 equals log subscript 10 space 16 times log subscript c space 10$ where $c greater than 0 semicolon c not equal to 1 semicolon b greater than 1 semicolon b not equal to 1$ determine b –

Just like we can change the base $b$ for the exponential function, we can also change the base $b$ for the logarithmic function. The logarithm with base $b$ is defined so that $log subscript b subscript blank end subscript c space equals space k$

If $log subscript c space 2 times log subscript b space 625 equals log subscript 10 space 16 times log subscript c space 10$ where $c greater than 0 semicolon c not equal to 1 semicolon b greater than 1 semicolon b not equal to 1$ determine b –

Statement 1:The two Fe atoms in $Fe subscript 3 straight O subscript 4$ have different oxidation numbers
Statement 2: $Fe to the power of 2 plus end exponent$ ions decolourise $KMnO subscript 4$ solution

$log subscript A space B$ , where $B equals fraction numerator 12 over denominator 3 plus square root of 5 plus square root of 8 end fraction$ and $A equals square root of 1 plus square root of 2 plus square root of 5 minus square root of 10$ is

In this question, we have to find the $log subscript a B$ . Here Solve first B. In B, rationalize it means multiply it denominator by sign change in both numerator and denominator

$log subscript A space B$ , where $B equals fraction numerator 12 over denominator 3 plus square root of 5 plus square root of 8 end fraction$ and $A equals square root of 1 plus square root of 2 plus square root of 5 minus square root of 10$ is

In this question, we have to find the $log subscript a B$ . Here Solve first B. In B, rationalize it means multiply it denominator by sign change in both numerator and denominator

Statement 1:The equivalence point refers the condition where equivalents of one species react with same number of equivalent of other species.
Statement 2:The end point of titration is exactly equal to equivalence point

maths-

If $alpha$ and $beta$ are the roots of the equation $open parentheses log subscript 2 space x close parentheses squared plus 4 open parentheses log subscript 2 space x close parentheses minus 1 equals 0$ then the value of $Error converting from MathML to accessible text.$ equals

maths-General

Statement 1:Iodimetric titration are redox titrations.
Statement 2:The iodine solution acts as an oxidant to reduce the reductant $straight I subscript 2 plus 2 straight e not stretchy rightwards arrow 2 straight I to the power of minus$

Statement 1:Diisopropyl ketone on reaction with isopropyl magnesium bromide followed by hydrolysis gives alcohol
Statement 2:Grignard reagent acts as a reducing agent

The first noble gas compound obtained was:

Given that $log subscript p space x equals alpha$ and $log subscript g space x equals beta$ then value of $log subscript p divided by q end subscript space x$ equals –

These four basic properties all follow directly from the fact that logs are exponents.
log_b(xy) = log_bx + log_by.
log_b(x/y) = log_bx - log_by.
log_b(xⁿ) = n log_bx.
log_bx = log_ax / log_ab.

Given that $log subscript p space x equals alpha$ and $log subscript g space x equals beta$ then value of $log subscript p divided by q end subscript space x$ equals –

These four basic properties all follow directly from the fact that logs are exponents.
log_b(xy) = log_bx + log_by.
log_b(x/y) = log_bx - log_by.
log_b(xⁿ) = n log_bx.
log_bx = log_ax / log_ab.

Statement 1:Change in colour of acidic solution of potassium dichromate by breath is used to test drunk drivers.
Statement 2:Change in colour is due to the complexation of alcohol with potassium dichromate.

Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2 $straight pi$ ] is equal to

In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.

Total number of solutions of sin{x} = cos{x}, where {.} denotes the fractional part, in [0, 2 $straight pi$ ] is equal to

In this question, we have to find the number of solutions. Here we have fractional part of x. It is {x} and {x}= x-[x]. The {x} is always belongs to 0 to 1.

Statement 1:If a strong acid is added to a solution of potassium chromate it changes itscolour from yellow to orange.
Statement 2:The colour change is due to the oxidation of potassium chromate.