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Xelga [282]
3 years ago
9

There are 8 circles and 6 squares. what is the simplest ratio of squares to circles?

Mathematics
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

4:3

Step-by-step explanation:

Given function:

  • 8 circle by 6 squares

Divide similar factor by each term:

  • 8=4·2
  • 6=3·2

Rewrite:

  • 4:3

Therefore, the simplest ratio for 8 circles and 6 squares would be 4:3..

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A bar graph is a chart that uses bars to show comparisons of different types of data. The bars can either be horizontal or vertical. They are always straight if I’m not mistaken
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3 years ago
Evaluate the function f(x) = 2x3 - 6x + 1 for f(-2).
erma4kov [3.2K]

Answer:

- 3

Step-by-step explanation:

f(x) = 2 {x}^{3}  - 6x + 1 \\ plugging \: x =  - 2 \\ f( - 2) = 2( { - 2})^{3}  - 6( - 2)+ 1  \\  = 2 \times ( - 8) + 12 + 1 \\  =  - 16 + 13 \\  =  - 3 \\ f( - 2) =  - 3

7 0
3 years ago
Given that, 1 Chinese yuan = £0.13, convert £314 into Chinese yuan. Give
Maurinko [17]

Answer:

2415 Chinese yuan (rounded off to nearest yuan)

Step-by-step explanation:

£0.13 = 1 Chinese yuan

£314 = \frac{314}{0.13}×1 Chinese yuan

       = 2415 Chinese yuan (rounded off to nearest yuan)

5 0
3 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
I need help to figure out this question
ololo11 [35]

I think it would be 16.95 Hope this help you in someway.

8 0
3 years ago
Read 2 more answers
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