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3 years ago

Explain how a number line can be used to find the difference for 34-28

2 answers:

8 0

Take a piece of paper and write a line with dashes having a number below it until 50. Then Go to 34 on your number line and then go backwards by 28. (34 - 28) the answer should be 6. The number line helps you count steps forward/backwards. :)

<3 Aleah

Liula [17]3 years ago

3 0

A number line can be used to find the difference between 34-28 because you can count each space in-between each number (6).

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To eliminate the y terms and solve for x in the fewest steps, by which constants should the equation need multiplied by before a

ivanzaharov [21]

A. Multiplied by 2 for the first then for the second multiply by three.

Explanation: the equations already have opposite signs for y so reversing the signs would be counterproductive. (Eliminates B and D) and the easiest way is to get the numbers before the Y variable to be the same. The top by 2 and bottom by 3 would do that, leaving only A.

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2 years ago

bob pays $20 to have the internet installed and $15 a month for service. how much will bob pay for 1 year of internet use? a.$18

Genrish500 [490]

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3 years ago

Read 2 more answers

PLZ HELP QUICKLY , GET BRAINLY

Leviafan [203]

Answer:

D.) Fixed costs do not change no matter how much a business produces; variable costs do change.

Step-by-step explanation:

A variable cost varies with the amount produced, while a fixed cost remains the same no matter how much output a company produces.

I'm 100% sure that this is the answer.

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3 years ago

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

L'(t) = 34.416*e^(-0.18*t)

L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

6 cm/yr = 34.416*e^(-0.18*t)

e^(0.18*t) = 34.416 / 6

0.18*t = Ln(34.416/6)

t = Ln(34.416/6) / 0.18

t = 10 year

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3 years ago

Find the greatest common factor of the following monomials: 20a3 and 8a2

dimulka [17.4K]

GCF of given monomials are $4a^2$

Solution:

Given that we have to find the greatest common factor

Given monomials are:

$20a^3 \text{ and } 8a^2$

When we find all the factors of two or more numbers, and some factors are the same, then the largest of those common factors is the Greatest Common Factor

Let us first find the GCF of 20 and 8 and then find GCF of variables and then multiply them together

GCF of 20 and 8:

The factors of 8 are: 1, 2, 4, 8

The factors of 20 are: 1, 2, 4, 5, 10, 20

Then the greatest common factor is 4

$GCF\ of\ a^3 \text{ and } a^2\\\\a^3 = a^2 \times a\\\\a^2 = a^2$