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2 years ago

Now heres the tricky part

Mathematics

1 answer:

loris [4]2 years ago

4 0

13. Answer: n ≥ 108

Step-by-step explanation:

18 ≤ n ÷ 6

×6 ×6

108 ≤ n → n ≥ 108

Note: The line under the inequality symbol represents a closed dot

Graph: 108 ·---------------→

************************************************************************************

14. Answer: x > -4

Step-by-step explanation:

-3x < 12

÷(-3) ÷(-3)

x > -4 Flip the inequality sign when × or ÷ by a negative

Note: No line under the inequality symbol represents an open dot

Graph: -4 o-----------→

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The flagpole is 20 ft high and casts an 8 foot shadow, the oak tree casts a 12 foot shadow. How tall is the oak tree

Fudgin [204]

Answer: 4.8 ft

To answer this question you need to know how much shadow: actual height ratio. Flagpole is having 8 ft shadow with 20 ft actual height. The ratio should be= 8 ft: 20 ft= 0.4

Then multiply the ratio with the oak tree shadow. The equation would be:
0.4 x 12 ft= 4.8 ft

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3 years ago

Write √ 3 x √ 6 in the form b √2 where b is an integer.

tigry1 [53]

Answer:

Step-by-step explanation:

The catch here is to break up sqrt(6) into 2 parts that use primes to define sqrt(6)

sqrt(3)* sqrt(6)

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2 years ago

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3 years ago

A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria

Kaylis [27]

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$ .

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$ , we get

$\frac{dP}{dt} = kP$ .

Notice that this is a separable equation, so

$\frac{dP}{P} = kdt$ .

Then, integrating in both sides of the equality:

$\int\frac{dP}{P} = \int kdt$ .

We have,

$\ln P = kt+C$ .

Now, taking exponential

$P(t) = Ce^{kt}$ .

The next step is to find the value for the constant $C$ . We do this using the initial condition $P(0)=400$ . Recall that this is the initial population of bacteria. So,