Solve for x. x^3=1/27

Becca and Kimberly are running at a constant speed in a marathon. Becca runs at 4.5 miles per hour. Kimberly’s progress is shown

Georgia [21]

Answer:

Part A: Kimberly

Part B:

Becca = 22.5 miles, Kimberly = 25 miles

Step-by-step explanation:

Part A:

Given that Becca runs at a constant speed of 4.5 miles per hour, use the table to find the constant speed that Kimberly runs, and compare who runs faster.

Let x = time

y = distance

k = speed = y/x

Equation for each person can be written as: y = kx

Therefore:

✔️Equation for Becca if k = 4.5

y = 4.5x

✔️Find k (speed) of Kimberly using (2, 10):

Speed (k) = y/x

k = 10/2

k = 5 miles per hour

Equation for Kimberly would be:

y = 5x

Comparing their speed, Kimberly runs faster because she covers more miles per hour than Becca does.

Part B:

For Becca, substitute x = 5 into Becca's equation, y = 4.5x

Thus:

y = 4.5*5 = 22.5 miles

For Kimberly, substitute x = 5 into Kimberly's equation, y = 5x

Thus:

y = 5*5 = 25 miles

5 0

2 years ago

Complete the like terms to create an equvalent expression: -4r - 2r + 5

Marat540 [252]

-6r+5
the two like terms are the ones with R. combine -4r and -2r to get -6r

3 0

2 years ago

Find the sum of -6/ab + a^2/b^2

Ymorist [56]

Answer:

$\frac{1}{ab^{2} }$ (a³ - 6b)

Step-by-step explanation:

$\frac{-6}{ab} + \frac{a^{2} }{b^{2} }$

= $\frac{-6b + a^{3} }{ab^{2} }$

= $\frac{1}{ab^{2} }$ (a³ - 6b)

4 0

3 years ago

What are the possible values of x and y for two distinct points, (5, –2) and (x, y), to represent a function?

bonufazy [111]

The correct answer is:

Any real number for x except 5, and any real number for y.

Explanation:

A function is a relation in which each element of the domain, or x-value, is mapped to one element of the range, or y-value. This means that no x can be mapped to more than one y, so if we have the same number used twice for x, we do not have a function.

This means that 5 cannot be used for x in the other ordered pair.

Since there is no restriction on y in order to be a function, y can be any real number.

8 0

3 years ago

Read 2 more answers

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .

Verify that $(x + 2)\, (x + 3)$ indeed expands to the original polynomial:

$\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}$ .

4 0

2 years ago