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2 years ago

The sum of eight times a number and negative four exceeds twelve.

Mathematics

2 answers:

Ganezh [65]2 years ago

5 0

Answer: Our required inequality would be

$(8x-4)>12$

Step-by-step explanation:

Let the number be 'x'.

Eight times a number is expressed as

$8x$

Sum of eight times a number and negative four is expressed as

$8x-4$

the above exceeds twelve is expressed as

$(8x-4)>12$

Hence, our required inequality would be

$(8x-4)>12$

Aliun [14]2 years ago

3 0

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Answer:

$92

Step-by-step explanation:

add what you saved and the parents gave you
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1 year ago

what is 1608 divided by 268, my answer was 5 but apparently its 6 so please explain why and show work :)

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2 years ago

Gino has a stick of pepperoni that is 26 1/2 inches long. He wants to cut 1/2 inch pieces to put on his large pizza. How many pi

ikadub [295]

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Step-by-step explanation:

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2 years ago

Weekly demand for a particular item averages 30 units, with a standard deviation of 4. This item is managed with a fixed-order-i

aivan3 [116]

Answer:

(B)93

Explanation:

Since we are using a fixed-order-interval model,

The Amount to Order=Expected Demand During protection Interval+Safety Stock-Amount at Hand

$=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A$

Where:

d=weekly demand

OI=Order Interval

LT=Lead Time

z=Standard Deviation of Desired Service Level

$\sigma_{d}$ =Standard Deviation of weekly Demand

A= Amount at Hand

$=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A$

$[30(3 + 1)] + [1.964*4*\sqrt{3 + 1} - 43$

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4 0

1 year ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago