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2 years ago

The graph represents function 1 and the equation represents function 2:

Mathematics

1 answer:

Nookie1986 [14]2 years ago

6 0

Answer:

Step-by-step explanation:

sikeee;( be mad

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Factor the polynomial below.<br> 9x squared minus 16y squared

GaryK [48]

9x^2-16y^2=
(3x+4y)(3x-4y)

7 0

3 years ago

Choose the slope-intercept form of 3x + 2y = 5.

Nataly_w [17]

Answer:

y = -3x/2 + 5/2

Step-by-step explanation:

Well, basically what you do is modify it so that y is on one side.

3x+2y = 5

2y = 5-3x

y = (5-3x)/2

y = 5/2 -3x/2

So, the answer is the second option.

4 0

3 years ago

Read 2 more answers

If you went to the hospital and you have a deductible of $250 and your bill was $1,000 dollars, how much money do you owe?

MA_775_DIABLO [31]

Answer:

The money which will be owe is $750

Step-by-step explanation:

Given as :

The hospital bill which is charge = $1000

The deductible amount in hospital bill = $250

So, The rest amount will be owned

∴ The Amount which will be owned = Total bill - Deductible bill

Or,The Amount which will be owned = $1000 - $250 = $750

Hence The Amount which will be owned is $750 Answer

4 0

3 years ago

What is the answer to this question?

Charra [1.4K]

Answer:

im just in 6 grade

Step-by-step explanation:

6 0

2 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