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lesantik [10]
3 years ago
12

IM ALMOST DONE WITH MY DIAGNOSTIC YAY!!!!​

Mathematics
2 answers:
Harman [31]3 years ago
8 0
The answer is 27units!! i think i am right
k0ka [10]3 years ago
4 0

Answer:

27 cubic units

hope this helps

have a good day :)

Step-by-step explanation:

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Hannah's school hosted a book donation. There are 150 students at her school, and they donated a total of b books! Hannah donate
svetoff [14.1K]

Answer:

x=3*\dfrac{b}{150}

Step-by-step explanation:

Let x = number of books that Hannah donated

We know that the number of students is 150 and the total number of books is represented by b, we can write that as:

Average number of books donated by each student = \dfrac{b}{150}

Now we know that Hannah donated 3 times as much as the average. We can write that as:

x=3*\dfrac{b}{150}

4 0
3 years ago
Solve the equation for an ellipse for y. Assume that y > 0. y^2/a^2 + x^2/b^2 = 1
In-s [12.5K]
Multiply each term by a^2b^2:-

b^2y^2 + a^2x^2 = a^2b^2
subtract a^2x^2 from both sides
b^2y^2  = a^2b^2  - a^2x^2
Now divide both sides  by b^2

y^2 =  a^2 -  a^2x^2 / b^2  = a^2 (1 - x^2/b^2)

take positive square root ( because y > 0)

y  = a sqrt(1 - x^2/b^2)
4 0
4 years ago
Read 2 more answers
Find the coordinate of K' after a Glide reflection of the triangle: translation 4 units up and 4 units right then a reflection a
monitta

Answer:

  a.  (-1, 3)

Step-by-step explanation:

The "glide" 4 units right and up is the transformation ...

  (x, y) ⇒ (x+4, y+4)

The reflection across y=1 is the transformation ...

  (x, y) ⇒ (x, 2-y)

Combined, you have the transformation ...

  (x, y) ⇒ (x +4, 2-(y +4)) = (x +4, -2-y)

Then, ...

  K(-5, -5) ⇒ K'(-5 +4, -2-(-5)) = K'(-1, 3)

4 0
3 years ago
Answers for question 17 and 18?
Tom [10]

Answer: think the answers near tenth

6 0
4 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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