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3 years ago

Y=2 + x; use x= 1, and y= 2

Mathematics

1 answer:

torisob [31]3 years ago

6 0

Y=2+1
2+1=3
So y=3
The answer is y=3

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8 3/4 +n+ 2 3/8 +23 1/8 solve for n

bezimeni [28]

8 3/4+n+2 3/8+ 23 1/8
8 3/4+n+ 25 4/8
35/4+ n+ 204/8
70/8 +n+ 204/8
274/8 +n
n= -274/8= -25 1/2

4 0

3 years ago

The log of x cubed times y squared simplified

tresset_1 [31]

Answer:

$log(x^{3}y^{2}) = 3 log x+2 log y$

Step-by-step explanation:

Step 1:-

using logarithmic formula $log(ab)=log a+log b$

so given $log(x^{3} y^{2} ) = log (x^{3} )+log(y^{2} )$

now simplify

= $3 log x+2 log y$

Answer:-

$log(x^{3}y^{2})= [tex]3 log x+2 log y$

4 0

3 years ago

In a recent awards ceremony, the age of the winner for best actor was 35 and the age of the winner for best actress was 54. For

Arada [10]

Answer:

The actress has the more extreme age because

Step-by-step explanation:

Given

Male Athletes:

$x = 35$

$\=x = 42.1$

$SD = 6.6$

Female Athletes

$x = 54$

$\=x = 34.3$

$SD = 10.1$

Required

Determine which athlete had more extreme age

To do this, we simply calculate the standard z score using

$z = \frac{x - \=x}{SD}$

For the actor:

$z = \frac{35 - 42.1}{6.6}$

$z = \frac{-7.1}{6.6}$

$z = 1.0758$

For the actress:

$z = \frac{54 - 34.3}{10.1}$

$z = \frac{19.7}{10.1}$

$z = 1.950$

Comparing both z values;

The actress has the more extreme age because

It has a positive z value
Its z value is greater than that of the actor

6 0

3 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

3 0

3 years ago

Rearrange, P=2l+2w solve for w.

Gennadij [26K]

P= 2l+2w

P-2l= 2w

1/2P -l= w

Rearrange into standard form.

w= 1/2P -l

8 0

3 years ago