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3 years ago

How do you solve u/5+u/10-u/6=1

1 answer:

8 0

Find the least common denominator for each of these fractions then complete the equation until it equals 1.

Common factor is 30
u/5 = 6u/30
u/10 = 3u/30
-u/6 = -5u/30

6u/30 + 3u/30 - 5u/30 = 1
4u/30 = 1
4u = 30
u = 7.5

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A group of 5 students are running a relay race. If the race is 1/4 of a mile long and each student runs the same distance how fa

abruzzese [7]

1/20 mile. divide 1/4 by 5 by fliping the division sign and 5. (1/4)(1/5)=1/20

6 0

3 years ago

Read 2 more answers

If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction

valentina_108 [34]

Answer:

a) $\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$ .

b) $Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}$

Step-by-step explanation:

Given a function $f(x,y,z)$ , this function has the following gradient:

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

(a) find the gradient of f

We have that $f(x,y,z) = x\sin{yz}$ . So

$f_{x}(x,y,z) = \sin{yz}$

$f_{y}(x,y,z) = xz\cos{yz}$

$f_{z}(x,y,z) = xy \cos{yz}$ .

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

$\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}$ .

$\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)$

The vector is $v = i + 3j - k = (1,3,-1)$

To use v as an unitary vector, we divide each component of v by the norm of v.

$|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}$

$v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})$

Now, we can calculate the scalar product that is the directional derivative.

$Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}$

6 0

3 years ago

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an

Genrish500 [490]

Answer:

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

Standard deviation $(\sigma)$ = 7.1

Mean $\overline{(y)}$ = 110

$\alpha =$ 1 - confidence interval = 1 - .98 = .02

$z_{\frac{\alpha}{2}}$ = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = $\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}$