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3 years ago

51=3kbut what is k?tysm for everyone helping me:)

Mathematics

1 answer:

Kaylis [27]3 years ago

8 0

Answer:

Step-by-step explanation:

51÷3 = 17

k=17

not sure what else to add but my answer had to be longer

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AleksAgata [21]

Answer: 3 • ? = -6

Reasoning: -6/ 3 = -2
3 • -2 = -6

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2 years ago

Write an expression to represent: The sum of ten and the quotient of a number and 6.

hoa [83]

Answer:

10 + x/6

Step-by-step explanation:

8 0

3 years ago

use the identity below to complete the tasks a^3-b^3=(a-b)(a^2+ab+b^2) when using the identity for the difference of two cubes t

zheka24 [161]

Answer:

see explanation

Step-by-step explanation:

Given that the the difference of cubes is

a³ - b³ = (a - b)(a² + ab + b²)

Given

64 $x^{6}$ - 27 ← a difference of cubes

with a = 4x² and b = 3, thus

= (4x²)³ - 3³

= (4x² - 3)(16 $x^{4}$ + 12x² + 9) ← in factored form

7 0

3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

Given:

$y=\left(x+\sqrt{1+x^2}\right)^m$

First derivative

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

Second derivative

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

Proof

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

1 year ago

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.

marishachu [46]

Answer:

$\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

Step-by-step explanation:

We have been given the function

$f(x)=-2x^2+4x^3+3x^2+18$

From the rational zeros theorem, we have

$\text{Possible rational zeros}=\pm\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}}$

From the given function,

Leading coefficient = 2

Factors of 2 are 1,2

Constant term = 18

Factors of constant term = 1, 2, 3, 6, 9, 18

Hence, we have

$\text{Possible rational zeros}=\pm\frac{1,2,3,6,9,18}{1,2}\\\\\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

6 0

3 years ago