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3 years ago

What is the 3rd root of 125

Mathematics

1 answer:

Natalija [7]3 years ago

8 0

Answer:

its five

Step-by-step explanation:

You might be interested in

What are the potential solutions to the equation 2ln(x+3)=0

skad [1K]

$2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1$

Expanding the left side gives

$x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0$

which gives two solutions, $x=-4$ and $x=-2$ . But if $x=-4$ , then $\ln(x+3)=\ln(-1)$ , but this number isn't real, so $x=-4$ is an extraneous solution. Meanwhile if $x=-2$ , you get $\ln(-2+3)=\ln1=0$ , so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.

4 0

3 years ago

Read 2 more answers

Please help! I attached the question below.

kompoz [17]

Answer:

$\frac{2(c+2)}{c(c-2)}$

Step-by-step explanation:

$\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }$

Identity used:

$a^{2}-b^{2}=(a-b)(a+b)$

$\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }$

Now let us divide the modified expressions:

$\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})}$ ÷ $\frac{(c-2)^2}{2(6c^{3}+15c^{2}) }$

we get:

$\frac{2(c+2)}{c(c-2)}$

5 0

3 years ago

How do you do this question?

vaieri [72.5K]

Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red plus (+) signs indicate the regions in which the function is concave up, and the red negative (-) signs indicate the regions in which the function is concave down.

Note that the sign of the concavity changes at an inflection point.

Let's examine the intervals given.

(-∞, -2): Yes, concave up.

(-∞, -1.17): No. Concave up in (-∞, -2) but concave down in (-2, -1.17).

(-2, 0): No. Concave down in (-2, -1.17) but increasing in (-1.17, 0.0).

(-1.17, 0.689): Yes. Concave up.

(-0.618, 1.618): No. Concave up in (-0.618, 0.689) but concave down in (0.689, 1.618).

(0, 3): No. Concave up in (0, 0.689), concave down in (0.689, 2.481), and concave up in (2.481, 3).

(2.481, ∞): Yes. Concave up.

The three intervals that are concave up are (-∞, -2), (-1.17, -0.689), and (2.481, ∞).

4 0

3 years ago

Which of the equations below could be the equation of this parabola?

nirvana33 [79]

Answer:

$y=-4x^2$ is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

$y=-4x^2$

$\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

$\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}$

$\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)$

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)$

$\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}$

$y=-4x^2$

$\mathrm{The\:parabola\:params\:are:}$

$a=-4,\:m=0,\:n=0$

$x_v=\frac{m+n}{2}$

$x_v=\frac{0+0}{2}$

$x_v=0$

$\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-4\cdot \:0^2$

$y_v=0$

Therefore, the parabola vertex is

$\left(0,\:0\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

$a=-4$

$\mathrm{Maximum}\space\left(0,\:0\right)$

so,

$\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)$

Therefore, $y=-4x^2$ is the equation of this parabola. The graph is also attached.

7 0

3 years ago

The function d(s) = 0.0056s squared + 0.14s models the stopping distance

victus00 [196]

Answer:

The car must have a speed of 25 kilometres per hour to stop after moving 7 metres.

Step-by-step explanation:

Let be $d(s) = 0.0056\cdot s^{2} + 0.14\cdot s$ , where $d$ is the stopping distance measured in metres and $s$ is the speed measured in kilometres per hour. The second-order polynomial is drawn with the help of a graphing tool and whose outcome is presented below as attachment.

The procedure to find the speed related to the given stopping distance is described below:

1) Construct the graph of $d(s)$ .

2) Add the function $d = 7\,m$ .

3) The point of intersection between both curves contains the speed related to given stopping distance.

In consequence, the car must have a speed of 25 kilometres per hour to stop after moving 7 metres.

4 0

3 years ago