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alexdok [17]
3 years ago
8

Dani leaves her house at 08 00.

Mathematics
1 answer:
leonid [27]3 years ago
3 0

Answer:

wth

Step-by-step explanation:

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A+0=0+a=0 is true except for all teal numbers except 0 true or false
jolli1 [7]

Try it with some random number, like a=3.

       Is "3+0 = 0+3 = 0" true?    No.

If you were doing multiplication, then it would be true, but not with addition.

6 0
3 years ago
Describe a sequence of transformations that can be performed to polygon ABCDE to prove that it is similar to polygon FGHIJ
maxonik [38]

Answer:

answer below

Step-by-step explanation:

ABCDE go through dilation over center (6 , -2) with factor of 1/2 to FGHIJ

AB // FG slope: -2 , √20:√5 = 2: 1

BC // GH // X axis 8:4 = 2:1

CD // HI, slope= 1 , √8:√2 = 2:1

DE // IJ // x axis, 4:2 = 2:1

EA // JF // y axis, 2:1

8 0
3 years ago
Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,
rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t ,  y_2(t) = 0   \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1 $ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t  \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0,  ⇒ dx = 0

y = 9 t,  ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$ \oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx  $

                        = 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy $

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

                            $= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$  is conservative.

5 0
3 years ago
What is a tenth of 12,000
almond37 [142]
To answer this we need to divide 12,000 by 10 and our answer will be the tenth of 12,000. Then we will check our work. Lets do it:-

12,000 ÷ 10 = 1,200
1,200 is the tenth of 12,000.

CHECK OUR WORK:-

1,200 × 10 = 12,000
We were RIGHT!!

So, 1,200 is the tenth of 12,000.

Hope I helped ya!! 
3 0
4 years ago
Read 2 more answers
2x^2 + 9x +12+3x+2 8x^2 write an equivalent expression.
Alex73 [517]

Answer:

10x^2 + 12x + 14

Step-by-step explanation:

Combine like terms: 2x^2 + 8x^2 = 10x^2 , 9x + 3x = 12x , 12 + 2 = 14

In total: 10x^2 + 12x + 14

7 0
3 years ago
Read 2 more answers
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