Which ordered pairs are a solution to y=4x?

Marizza181 [45]

Answer:

$x^{5/6 } + 2\cdot x^{7/3}$

Step-by-step explanation:

Let be $x^{1/3}\cdot (x^{1/2} + 2\cdot x^{2})$ , which can be simplified solely by algebraic means. We proceed to describe all the steps needed for simplification:

1) $x^{1/3}\cdot (x^{1/2} + 2\cdot x^{2})$ Given.

2) $x^{1/3}\cdot x^{1/2} + 2\cdot (x^{1/3}\cdot x^{2})$ Distributive and associative properties.

3) $x^{5/6 } + 2\cdot x^{7/3}$ $a^{m}\cdot a^{n} = a^{m+n}$ /Result

6 0

2 years ago

What is the answer?<br> 2500/x+20<br> =2500/(48.59)+20

sammy [17]

The answer would be 71.45

7 0

3 years ago

What is the point used in the equation of the line y+4=1/2(x-2)

Ket [755]

Answer:

What is the point used in the equation of the line y+4=1/2(x-2)

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:

y - y1 = m(x - x1)

Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula

They've given me m = 4, x1 = -1, and y1 = -6. I'll plug these values into the point-slope form, and solve for "y=":

y - y1 = m(x - x1)

y - (-6) = (4)(x - (-1))

y + 6 = 4(x + 1)

y + 6 = 4x + 4

y = 4x + 4 - 6

y = 4x - 2

8 0

3 years ago

I need help ASAP?!!!!!!

Elodia [21]

Answer:

x = 2

Step-by-step explanation:

5x - 6 = 4

add 6 to each side

5x = 10

divide each side by 5

x = 2

3 0

2 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

3 years ago