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Deffense [45]
2 years ago
12

Brad took 5 kicks and made 4 goals: 4/5

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

Answer: 8/10

Step-by-step explanation: so you have to multiply 5 times x u till you get 10 so you’ll multiply it 2 times and since you multiplied the left 5 times two you have to multiply the 4 by 2 also to keep everything equal so the final answer is 8/10

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Is it possible to make a triangle with sides lengths of 12, 16 and 20? If the answer is yes, is the triangle a right triangle?
Sunny_sXe [5.5K]

Answer: Yes it can form a triangle and its a right triangle.

Step-by-step explanation:

Using the Pythagorean theorem we can see that the longest side (20) is our hypotenuse. From there you just check if its true. :)

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
2) Ten Percent of 27 is _<br> One Percent of 27 is<br> Find 14% of 27.
geniusboy [140]

Step-by-step explanation:

10%of 27

=  \frac{10}{100}  \times 27

=  \frac{27}{10}  = 2.7

1% of 27

=  \frac{1}{100}  \times 27 = 0.27

14\%of \: 27 =  \frac{14}{100}  \times 27 = 3.78

8 0
4 years ago
What is 1/3 x 3/2 equal
mel-nik [20]

Answer:

3/6

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The cost of a phone is reduced by 20%. The new cost is $70.40. What was the original price?
Anton [14]

Answer:

.8x = $70.40

x = 70.4/.8

x = $88

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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