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masya89 [10]
2 years ago
13

Whats 2-2? I'm 8 Please help me

Mathematics
1 answer:
jolli1 [7]2 years ago
6 0

Answer:

0

Step-by-step explanation:

2-2=0

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What is the slope of the function?<br> -10<br> -5<br> 5<br> 10
bixtya [17]

Answer:

m= 1/2

Step-by-step explanation:

3 0
3 years ago
How do i solve 42 = -2d + 6
Mazyrski [523]
Substract 6 to both sides and you get 36 =-2d then you use the multiplicative inverse and d=-18.
8 0
3 years ago
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What is the value of y?<br> 12<br> 13<br> х
krek1111 [17]

Answer:

y is 6.5 because it's kind of a ratio, you just need to find out what number to divide them by or multiply them by in this case, it was 13÷2 =6.5

3 0
3 years ago
If g(x) = 5x2^2 - 10x + 9 and h(x) = -3x3^3+ 5x - 6, find g(x) - h(x)?
AnnZ [28]

Answer:

B)g(x) - h(x) =  3x^3 - 5x^2 - 15x + 15

Step-by-step explanation:

Here, the given functions are:

g(x) = 5x^2- 10x + 9 \\h(x) = -3x^3+ 5x - 6

Now, g(x) - h(x) is : 

(5x^2- 10x + 9)  - ( -3x^3+ 5x - 6)\\= 5x^2- 10x + 9 + 3x^3 - 5x + 6\\= 3x ^3 + 5x ^2+  (-10 x - 5 x) + (9 + 6)\\= 3x^3 - 5x^2 - 15x + 15

⇒(5x^2- 10x + 9)  - ( -3x^3+ 5x - 6) =  3x^3 - 5x^2 - 15x + 15

B) Hence, g(x) - h(x) =  3x^3 - 5x^2 - 15x + 15

3 0
2 years ago
Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2
LUCKY_DIMON [66]
Make a change of coordinates:

u(x,y)=xy
v(x,y)=\dfrac xy

The Jacobian for this transformation is

\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}

and has a determinant of

\det\mathbf J=-\dfrac{2x}y

Note that we need to use the Jacobian in the other direction; that is, we've computed

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}

but we need the Jacobian determinant for the reverse transformation (from (x,y) to (u,v). To do this, notice that

\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to

\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv
=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2
8 0
3 years ago
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