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sergey [27]
3 years ago
6

What is the range of ƒ(x) = c?

Mathematics
1 answer:
nikdorinn [45]3 years ago
3 0
All real numbers because it’s a function that has no limit
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Which equation shows y=−12x+4 in standard form?
Tema [17]
The answer is the first option

6 0
2 years ago
9n - 15 for n = 5 help me please
MaRussiya [10]

If n = 5, then 9n = 45

45 - 15 = 30, which is the same thing as 6n


7 0
3 years ago
Which of the following statements is true?
Vitek1552 [10]

Answer:

B

Step-by-step explanation:

1/9 can be simplified to 1/3 as 1/3 x 1/3 = 1/9  and 49/25 can be simplified to 7/5 as 7/5 x 7/5 = 49/25.

Now looking at both the fraction 1/3 and 7/5

1/3 simplifies to 0.34

It is an irrational value as the 3 keeps repeating.

7/5 simplifies to 1.4

It is a rational value as the decimal value is not repeating.

Thus making the second option the answer.

6 0
2 years ago
2x (3x5) - (8-3)= what is the answer please
bekas [8.4K]

Answer:

2x15-5=30-5=25 So, the answer is 25.

6 0
2 years ago
Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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