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2 years ago

What is the answer? Simplify (v + 6) 3

Mathematics

2 answers:

Anika [276]2 years ago

5 0

3v+18 :) that’s it!!

Allisa [31]2 years ago

3 0

Answer:

<h2>The answer will be 3v+18</h2>

Step-by-step explanation:

<h2>Brainliest please?</h2>

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Two students went to the schools clinic with fevers Isabell had a temperature of 103.78 and Bryan had a temperature that was 2.3

jeyben [28]

Answer:

1) The temperature of Bryan is 101.39°

Step-by-step explanation:

Here, the Temperature of Isabell is given = 103.78 degrees

Now, the temperature of Bryan = Iaabell's Tempt. - 2.36

So, temperature of Bryan = (103.78 - 2.36) degrees

= 101.4 degree

Hence, the temperature of Bryan = 101.4 degree

So, the most appropriate option is 101.39°

4 0

2 years ago

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Ivenika [448]

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${x}^{2} - {y}^{2} + 4y - 4 \\ \\ {x}^{2} - ( {y}^{2} .2.y.y + {2}^{2} ) \\ \\ {x}^{2} - (y - 2) {}^{2} \\ \\ (x + y - 2)(x - y + 2)$

4 0

2 years ago

Read 2 more answers

Last math question.

siniylev [52]

Answer:

Step-by-step explanation:

$\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-\text{slope}\\b-\text{y-intercept}\\\\\text{Let}\\\\k:y=m_1x+b_1\\l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\k\ ||\ l\iff m_1=m_2\\=========================$

$\text{We have}\ y=5x+2\to m_1=5\\\\\text{therefore}\ m_2=5.\\\\\text{Put the value of the slope and the coordinates of the given point (-2, 1)}\\\text{to the equation of a line:}\\\\1=5(-2)+b\\1=-10+b\qquad\text{add 10 to both sides}\\1+10=-10+10+b\\11=b\to b=11\\\\\text{Finally:}\\\\y=5x+11$

5 0

3 years ago

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:

<u>Given</u>:

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

<u>First derivative</u>

$\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)$\\\end{minipage}}$

<u />

<u /> $\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}}$

<u />

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

<u>Second derivative</u>

<u />

$\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}$\\\end{minipage}}$

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

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

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

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

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

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

<u>Proof</u>

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

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

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

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

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

$= 0$

8 0

1 year ago

PLEASE HELP KINGSSSS

creativ13 [48]

Answer:

30/52

Step-by-step explanation:

its a 26/52% chance to get a black card because 13 spain cards and 13 clovers add it together you get 26 and theirs 4 kings in a dack so add them both together 26/52+4/52=30/52

<h2>Answer: 30/52</h2>

3 0

2 years ago