Ask question

Ask question

All categories

3 years ago

8

If 12 assemblers can compleate a certain job in 6 hours, how long will the same job take if the number of assemblers is cut back

to 9?

1 answer:

marshall27 [118]3 years ago

3 0

Answer:

8 hours

Step-by-step explanation:

total labor hours required

12 * 6 = 72

hours per worker with 9 workers

72/9 = 8

8 hours

You might be interested in

Which choice is equivalent to the product below when x>0?

V125BC [204]

Answer:

D

Step-by-step explanation:

Using the rule of radicals

$\sqrt{\frac{a}{b} }$ = $\frac{\sqrt{a} }{\sqrt{b} }$

$\sqrt{a\\}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Given

$\sqrt{\frac{6}{x} }$ × $\sqrt{\frac{x^2}{24} }$

= $\frac{\sqrt{6} }{\sqrt{x} }$ × $\frac{x^2}{24}$

= $\frac{\sqrt{6} }{\sqrt{x} }$ × $\frac{x}{2 \sqrt{6\\} }$

Cancel $\sqrt{6}$ on numerator/ denominator

= $\frac{1}{\sqrt{x} }$ × $\frac{x}{2\\}$

= $\frac{1}{\sqrt{x} }$ × $\frac{(\sqrt{x})^2 }{2}$

Cancel $\sqrt{x}$ on numerator/ denominator, leaving

= $\frac{\sqrt{x} }{2}$ → D

4 0

3 years ago

A cell phone company charges $400 for a new phone and $50 for a monthly plan. If C (t) is a rational function that represents th

guapka [62]

Answer:

its c

Step-by-step explanation:

7 0

3 years ago

George has a mass of 65kg. What would George weigh on the Moon? (The gravitational field strength on the Moon is 1.6N/kg.)

maks197457 [2]

George would weigh

65 kg × 1.6N /kg = 104 Newtons

That's compared to around 650 Newtons on earth.

3 0

3 years ago

Read 2 more answers

Javier and his study group designed a word problem, equation, table, and graph that were all supposed to represent the same info

yKpoI14uk [10]

Answer:

Graph

Step-by-step explanation:

for 5 tickets it would cost $40 but on the graph it shows $1.

8 0

3 years ago

Read 2 more answers

The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π

Gennadij [26K]

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

or, via symmetry

$\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

____________

(b)

By the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}$

For polar coordinates, x = rcosθ and y = rsinθ. Since
<span>r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ 
y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

$\begin{aligned}
\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\
&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\
\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}
&= 2/3
\end{aligned}$

2/3 is the slope

____________

(c)

"</span><span>distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time

</span><span /> $\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$

5 0

3 years ago