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2 years ago

Which set of ordered pairs does not represent a function?

Mathematics

1 answer:

Pachacha [2.7K]2 years ago

8 0

Answer:

{(8,−4),(−5,6),(−5,7),(−3,−5)}

Step-by-step explanation:

because for any domain ,you will get only one range value ,

but ,there are two range values for -5 in A;

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The height of a cylinder is 10 and the area of a base is 36 pi square units. What

ollegr [7]

Answer:

1130.97336 units^3

Step-by-step explanation:

The volume of a cylinder can be found using:

$v=\pi r^2h$

We have the area of the base, but not the radius

$a=\pi r^2$

We know the area is $36\pi$ , so we can substitute that in for a

$36\pi =\pi r^2$

We want to find r, so we need to isolate it

Divide both sides by pi

36=r^2

Take the square root of both sides

6=r

Now we know the radius, and can substitute it into the volume formula, and we can substitute the height (10) in

$v=\pi r^2h$

$v=\pi 6^210$

Solve the exponent

$v=\pi 36(10)$

$v=\pi 360$

v=1130.97336

The volume is 1130.97336 units^3

6 0

3 years ago

What is the square root of 9

Troyanec [42]

The square root of 9 is 3

4 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Let T : V → W be a linear transformation from vector space V into vector space W.

wel

Explanation:

Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that

a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.

Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.

7 0

3 years ago

Translate to an equation and solve: The sum of two-fifths and f is one-half.<br> Provide<br> f=

SCORPION-xisa [38]

Answer:

9/10

1/2+2/5

Since these fractions have different denominators, we need to find the least common multiple of the denominators.The least common multiple of 2 and 5 is 10, so we need to multiply to make each of the denominators = 10

1/2 ∗ 5/5 = 5/10

5/2∗ 2/2= 4/10

Since these fractions have the same denominator, we can just add the numerators

5/10+ 4/10 = 9/10

3 0

2 years ago