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2 years ago

What is the transformation of f(x)=- (x + 2)2- 4

Mathematics

1 answer:

pychu [463]2 years ago

6 0

Answer: x=0

Step-by-step explanation: f(x)=(x+2)x2-4

0=2x+4-4

0=2x

2x/2=0

0/2=0

x=0

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Select ALL the correct answers.

jolli1 [7]

Answer:

(4x+5)(-3x-1) = -12x²-19x-5

Option A:

(-16x² + 10x - 3) + (4x² - 29x - 2) = -12x²-19x-5

Option A is correct.

Option B:

3(x - 5) - 2(6x² + 9x + 5) = -12x²-15x-25

Option B is wrong.

Option C:

2(x - 1) - 3(4x² + 7x + 1) = -12x²-19x-5

Option C is correct.

Option D:

(2x² - 11x - 9) - (14x² + 8x - 4) = -12x²-19x-5

Option D is correct.

4 0

2 years ago

Two times x is 8 more than y. The sum of x and two times y is 14. Write two equations and graph to find the value of y.

Sauron [17]

Option B: $y=4$ is the value of y

Explanation:

It is given that "Two times x is 8 more than y". Writing it as expression, we have,

$2x=8+y$

It is also given that "The sum of x and two times y is 14". Writing it as expression, we have,

$x+2y=14$

To find the value of y, let us solve the two equations using substitution method.

From the equation $2x=8+y$ , let us find the value of x.

$x=\frac{8+y}{2}$

$x=\frac{8}{2} +\frac{y}{2}$

$x=4+\frac{y}{2}$

Now, substituting $x=4+\frac{y}{2}$ in the equation $x+2y=14$ , we get,

$4+\frac{y}{2}+2y=14$

$4+\frac{5y}{2} =14$

$\frac{5 y}{2}=10$

$5y=20$

$y=4$

Thus, the value of y is $y=4$

8 0

3 years ago

What is the mean and median of the data<br><br> 4.7,8.51,6.5,7.42,9.64,7.2,9.3

Savatey [412]

Answer:Median=7.42

Mean=7.61

Step-by-step explanation:

4 0

2 years ago

Explain how you would convert a measurement given in ounces into pints

Greeley [361]

8 ounces = 1cup
2 cups = 1 pint
So, 16 ounces = 2 cups = 1pint
16 ounces = 1 pint

If you have 8 ounces, you would have half a pint.
If you have 20 ounces:
20 ounces/16 ounces= 5/4
So, you'd have 1 1/4 or 1.25 pints

4 0

3 years ago

If f(x) = f(x)g(x), where f and g have derivatives of all orders, show that f'' = f ''g + 2f 'g' + fg''.

Nadya [2.5K]

Differentiating once, we have

$f'(x)=f'(x)g(x)+f(x)g'(x)$

Differentiating again,

$f''(x)=f''(x)g(x)+f'(x)g'(x)+f'(x)g'(x)+f(x)g''(x)$
$f''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$

as needed.

5 0

3 years ago