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3 years ago

What is (2x)^3 I need help

2 answers:

8 0

So I think this is asking what’s 2x to the third power.
Well, that’s essentially saying what’s 2x Times 2x Times 2x

When we do that, we get 8x^3

Anastasy [175]3 years ago

3 0

Rewritten is (2x) (2x) (2x) which equals 8x^3

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Write a quadratic equation to represent a function with the vertex (-4, 1)

pishuonlain [190]

Answer:

y=a(x+4)^2 +1, for a real number a hat is different of 0.

Step-by-step explanation:

If we know only the vertex, there are many quandratic equations.

* Result: Vertex (m,n), the quadratic equation is y= a(x-m)^2 + n for a real number a that is different of 0.

Here, the vertex (-4,1), we have the equation: y=a(x+4)^2 +1, for a real number a hat is different of 0.

Hope you understand.

6 0

3 years ago

Please help i’ve been here for hours so can someone please tell me what to put here

andreev551 [17]

Answer:

If you divide the amount $60.20 (60.20) by 5 (amount of hours), you can find out the amount of money Andrew gets paid in dollars per hour. The quotient is 12.04, $12.04. Therefore, Andrew is payed $12.04 per hour.

Step-by-step explanation:

HOPE THIS HELPS

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6 0

3 years ago

Read 2 more answers

Find the length of the following curve. If you have a grapher, you may want to graph the curve to see what it looks like.

stepladder [879]

The length of the curve $y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6 is 192 units

<h3>How to determine the length of the curve?</h3>

The curve is given as:

$y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6

Start by differentiating the curve function

$y' = \frac 32 * \frac{1}{27}(9x^2 + 6)^\frac 12 * 18x$

Evaluate

$y' = x(9x^2 + 6)^\frac 12$

The length of the curve is calculated using:

$L =\int\limits^a_b {\sqrt{1 + y'^2}} \, dx$

This gives

$L =\int\limits^6_3 {\sqrt{1 + [x(9x^2 + 6)^\frac 12]^2}\ dx$

Expand

$L =\int\limits^6_3 {\sqrt{1 + x^2(9x^2 + 6)}\ dx$

This gives

$L =\int\limits^6_3 {\sqrt{9x^4 + 6x^2 + 1}\ dx$

Express as a perfect square

$L =\int\limits^6_3 {\sqrt{(3x^2 + 1)^2}\ dx$

Evaluate the exponent

$L =\int\limits^6_3 {3x^2 + 1} \ dx$

Differentiate

$L = x^3 + x|\limits^6_3$

Expand

L = (6³ + 6) - (3³ + 3)

Evaluate

L = 192

Hence, the length of the curve is 192 units