All categories

4 years ago

I need help were do I start

Mathematics

1 answer:

Reika [66]4 years ago

4 0

2. 4= 2², so neighboring perfect squares should look like 1² and 3²

1² 2² 3²

1 4 9

3. 16 = 4², so neighboring perfect squares are 3² and 5²

3² 4² 5²

9 16 25

4. 169 = 13²

12² 13² 14²

144 169 196

5. 9 = 3²

2² 3² 4²

4 9 16

6. 25 = 5²

4² 5² 6²

16 25 36

7. 49 = 7²

6² 7² 8²

36 49 64

You might be interested in

Which triangle drfined by the given points on the coordinate plane is similar to the triangle illustrated?

Harrizon [31]

Answer:D.) (-1, 1), (-1, 5), (-7, 1)

Step-by-step explanation:The length of the horizontal segment of the triangle graphed is 3. The length of the vertical segment of the triangle graphed is 2. Using the Pythagorean theorem, this makes the length of the missing side

2²+3²=c²

4+9 = c²

13=c²

√13 = c

This makes our side lengths 2, 3, and √13.

For choice A our vertices are (-1, 1), (-1, 4), (-6, 1).

This makes the length of the horizontal segment 5 and the length of the vertical segment 3. This makes the missing side length

5²+3²=c²

25+9=c²

34=c²

√34 = c

This is not proportional to the triangle drawn, so it is not similar.

The vertices of the triangle in option B are (-1, 1), (-1, 5), and (-6, 1). This makes the horizontal segment 5 units long and the vertical segment 4 units long. This makes the missing side length

4²+5²=c²

16+25=c²

41=c²

√41=c

These are not proportional to the triangle drawn, so the triangles are not similar.

The vertices of the triangle in choice C are (-1, 1), (-1, 4) and (-7, 1). This makes the length of the horizontal segment 6 and the length of the vertical segment 3. This makes the missing side length

3²+6²=c²

9+36=c²

45=c²

√45=c

3√5=c

These sides are not proportional to the sides of the triangle drawn, so the triangles are not similar.

The vertices of the triangle in choice D are (-1, 1), (-1, 5) and (-7, 1). This makes the length of the horizontal segment 6 and the length of the vertical segment 4. This makes the missing side length

6²+4²=c²

36+16=c²

52=c²

√52=c²

2√13=c

These sides are all twice as long as the sides of the triangle drawn; this means the triangles are similar.

4 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 :

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

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