What does -1 1/6 x 1/4 equal

What are co primes and primes Hrishiiiii pls help me

Anon25 [30]

Answer:

SEE BELOW

Step-by-step explanation:

Coprime numbers:

A Co-prime number is a set of numbers or integers which have only 1 as their common factor.

For eg: (2,3), (11,12), (99,100) and so on; they have 1 as their HCF. The sum of any two coprime numbers is always coprime with the product of the same two coprime numbers. For eg: The sum of 2 and 3 is 5 and the product of 2 and 3 is 6. Hence, 5 and 6 are coprime numbers.

Prime number

A number that can be divided exactly only by itself and 1, for example 7, 17 and 41

8 0

3 years ago

1)) An angle measures 132° less than the measure of its supplementary angle. What is the

Alexeev081 [22]

Answer:

156° and 24°

Step-by-step explanation:

Let it's supplementary angle be x.

So, the angle = (x - 132)°

As we know, sum of supplementary angles is 180°;

x + x - 132 = 180

2x = 180 + 132

2x = 312

x = 312/2

x = 156°

(x-132)° = (156-132)° = 24°

Hence, the angles are 156° and 24°.

5 0

3 years ago

Divide 4.16 into the ratio 3:5<br><br> need answers fast!!!

Over [174]

3:5 =3/5 3/5=4.16/x using cross multiplication, 5(4.15)=3(x) 20.8=3x x=108/15 thus, 4.16÷104/15 =3/5

4 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago

The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:

deff fn [24]

Step-by-step explanation:

Given equation of circle is,

${x}^{2} + {y}^{2} - 6y - 12 = 0 \\ by \: compairing \: it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f= - 6 \\ or \: f= - 3 \\ c = - 12$

again the another form of circle is,

${x}^{2} + {(y - k)}^{2} = 21 \\ or \: {x}^{2} + {y}^{2} - 2ky + {k}^{2} - 21 = 0 \\ by \: compairing \:it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\ g = 0 \\f = - k \\ c = {k}^{2} - 21$

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3

4 0

3 years ago