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3 years ago

Three different fraction equivalent 1

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

7 0

Answer:

3/3, 9/9, 12/12

Step-by-step explanation:

Equivilant fractions are only the ssame number on top of each other OR a number on top of a 1

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(-x + 3) + (x + 5) = i need a step by step explanation plz

ddd [48]

Answer:

(-x+3)+(x+5)= -x+3+x+5. =now add the numbers -x+8+x = now combined like terms = 8 now 8 is the answer

5 0

2 years ago

Suppose that 13 inches of wire cost 39 cent how much would 8 inches cost

Feliz [49]

Answer:

24 cent

Step-by-step explanation:

If 13 inches of wire cost 39 cent.

We have to calculate how much 1 inch of wire would cost first, before determining the cost of 8 inches of wire.

If 13 inches of wire costs 39cents, Then 1 inch of wire would cost 3 cent.

That is 39/13, which gives 3.

If 1 inch of wire cost 3 cents. Two inches of wire would cost 6 cents we 3 inches of wire costs 9 cents.

To get the cost of 8 inches of wire, we therefore multiply 8 by 3, which gives 24.

8 inches of wire costs 24 cents.

7 0

3 years ago

Is 0.963 close to 3/4

Alex777 [14]

There's no such concept as "close" in mathematics. Or at least, you have to specify when you consider two numbers to be "close".

All we can say is that, since 3/4=0.75, the two numbers are

$0.963-0.75=0.213$

units apart. Is this small enough to consider them as "close"? Is this big enough to consider them not to be "close"?

You should clarify more what you mean so that a definitive answer can be given.

4 0

2 years ago

Read 2 more answers

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}$

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago

Could someone help please

user100 [1]

Answer:

C, 3y(2y+9)

Step-by-step explanation:

you can factor out a 3 from both 6 and 27, as well as a y from both so you would put them on the outside as shown in C. Hope that helped! Let me know if you need more of an explanation, and id be happy to help!

6y^2+27y

6 and 27 both have a common factor, meaning that the same number can go in to them both, 3. Because both 6 and 27 have a y, you are able to also take out a y so outside would be 3y.

so the answer is 3y(2y+9)

4 0

3 years ago