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3 years ago

I don't understand this??

Mathematics

1 answer:

Sindrei [870]3 years ago

4 0

3,200?? (it may not be right I'm sorry.)

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What is cos 28 degrees

Diano4ka-milaya [45]

The trigonometric function gives the ratio of different sides of a right-angle triangle. The value of cos(28°) is -0.9626.

<h3>What are Trigonometric functions?</h3>

The trigonometric function gives the ratio of different sides of a right-angle triangle.

$\rm Sin \theta=\dfrac{Perpendicular}{Hypotenuse}\\\\\\Cos \theta=\dfrac{Base}{Hypotenuse}\\\\\\Tan \theta=\dfrac{Perpendicular}{Base}\\\\\\Cosec \theta=\dfrac{Hypotenuse}{Perpendicular}\\\\\\Sec \theta=\dfrac{Hypotenuse}{Base}\\\\\\Cot \theta=\dfrac{Base}{Perpendicular}\\\\\\$

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.

The cosine is the ratio of the base of the triangle and the hypotenuse of the triangle. Therefore, The value of cos(28°) is -0.9626.

Learn more about Trigonometric functions:

brainly.com/question/6904750

#SPJ1

6 0

2 years ago

Exit Ticket Anyone Can help me?

Umnica [9.8K]

Answer:

okay, so 1 1/6 can be converted to 7/6.

2 3/8 can be converted to 19/8

now find a common denominator, which is 24

7/6 would turn into 28/24

19/8 would turn into 57/24

now add! 28+57= 85/24

85/24

Step-by-step explanation:

please consider marking me as brainliest, my answer took alot of time and I am trying to rank up! have a great day lovely!!

4 0

3 years ago

Which matrix represents the system of equations shown below? 3x -7y=21

Serjik [45]

$\left\{\begin{array}{ccc}3x-7y=21\\4x+5y=11\end{array}\right \\\\.\ \ x\qquad y\ \ const.\\\left[\begin{array}{ccc}3&-7\\4&5\end{array}\right|\left\begin{array}{ccc}21\\11\end{array}\right]$

8 0

3 years ago

Which inequality can be used to represent a number x that is less than equal to 8 and greater than -3

KIM [24]

Answer: x = -2,-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

Solution:

A number line is used to represent the possible numbers that x can be.

x that is less than equal to 8 >>>> x ≤ 8

x is greater than -3 >>>> x > -3

So it will be represented as,

-3 < x ≤ 8

So all the numbers between -3 and 8 including eight (because it has the equal sign)

-2,-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

7 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:

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