All categories

3 years ago

I NEED SOMEONE TO HELP ASAP: I just need someone to classify all of these things

Mathematics

1 answer:

Sergio039 [100]3 years ago

7 0

Integer, natural, whole, rational, irrational, natural (15/3=5), integer (-12/6=-2), rational, irrational, rational, rational

Natural-the counting numbers (1,2,3,100...)
Whole-all natural AND 0
Integers-add negative numbers (no fractions)
Rational-add fractions and decimals
Irrational-numbers that can’t be a fraction (square root of 3, pi)
Real-all numbers

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Find f^-1(x) and it’s domain.

Mrac [35]

Answer: A

Step-by-step explanation:

Letting $f(y)=x$ ,

$x=\sqrt{y}-5\\\\x+5=\sqrt{y}\\\\y=(x+5)^{2}$

Also, the domain of an inverse is the same as the range of the original function, so the range is $x \geq 0$

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2 years ago

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3 years ago

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2 years ago

Consider the rule y = 5x + 7.

maksim [4K]

Answer:

see below

Step-by-step explanation:

Consider the rule y = 5x + 7.

a. How many tiles are in Figure 0?

In figure 0, we let x=0

y = 5*0+7 = 0+7 = 7

b. Which figure has 37 tiles?

This means y = 37

37 = 5x+7

Subtract 7 from each side

37-7 = 5x+7-7

30 = 5x

Divide by 5

30/5 = 5x/5

6 =x

Figure 6 has 37 tiles

c. In the equation y = mx + b, what do the letters m and b represent?

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3 years ago

Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

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3 years ago