Bryan played two baseball games last weekend. He got 7 hits in all.

Mathematics

2 answers:

Readme [11.4K]3 years ago

8 0

Answer:

4 hits in the second game

Step-by-step explanation:

If he hit 3 in the first game and 7 total, then the remainder (7-3) would be 4 for the second game,

Verdich [7]3 years ago

7 0

Answer: It's probably 4.

Step-by-step explanation:

7-3=4

You might be interested in

Y = -2x + 19 y = x + 7

andrey2020 [161]

System of equations:

$y=-2x+19$

$y=x+7$

The intersection point for this is (4,11).

3 0

4 years ago

Read 2 more answers

There is one error in one of five blocks of a program. To find the error, we test three randomly selected blocks. Let X be the n

marusya05 [52]

Probability(error in one block)= 1/5
Probability(error in 3 blocks)= 3*(1/5)= 0.6

Let,
0 = No error
 1 = error
Hence.
E(x) = 0.4 x 0 x+0.6 + 1
= 1.6

6 0

3 years ago

Read 2 more answers

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$