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2 years ago

Show all work to solve 3x^2 – 5x – 2 = 0.

Mathematics

2 answers:

scZoUnD [109]2 years ago

4 0

Answer:

I hope it will help

avanturin [10]2 years ago

3 0

Answer:

Step-by-step explanation:

3x2−5x−2=0

For this equation: a=3, b=-5, c=-2

3x2+−5x+−2=0

Step 1: Use quadratic formula with a=3, b=-5, c=-2.

x= (−b±√b2−4ac )2a

x= (−(−5)±√(−5)2−4(3)(−2) )/2(3)

x= (5±√49 )/6

x=2 or x= −1 /3

Answer:

x=2 or x= −1/ 3

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The following list gives the number of public libraries in each of 11 cities. 7, 11, 8, 7, 8, 7, 8, 11, 8, 10, 7 Find the modes

AlladinOne [14]

Answer:

The mode are: 7, 8

Step-by-step explanation:

Given

$7, 11, 8, 7, 8, 7, 8, 11, 8, 10, 7$

Required

Determine the mode

We start by arranging the given data in ascending order

The ordered data are:

$7,7,7,7, 8,8,8,8, 10, 11, 11$

From the above data.

7 has a frequency of 4

8 has a frequency of 4

10 has a frequency of 1

11 has a frequency of 2

The data with the highest frequency is the mode.

We can see that 7 and 8 both have frequencies of 4

Hence, the mode are: 7, 8

8 0

2 years ago

Read 2 more answers

15+(12+8a) simplify and justify each step

ki77a [65]

1. distribute a 1 to the 12 and 8a
2. you'll get 15+ 12 +8a
3. add the like terms(15 and 12)
4. your final answer will be 27+8a

5 0

2 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

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

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

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

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

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

$= 0$

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1 year ago

A professor has kept records on grades that students have earned in his class. If he wants to examine the percentage of students

AleksandrR [38]

Answer:

Pie chart

Step-by-step explanation:

Given that a professor has kept records on grades that students have earned in his class

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We have to find which kind of plot could he make

Box plot gives comparison for interval of scores this is to some extent helpful to find out the percentage of students getting a range of scores

Timeplot is irrelevant here. Dot plot will not be of much use.

Histogram is to some extent useful as box plot

But the best one is pie chart giving easy visual comparison for various grades A, B,C,D,E and F

8 0

2 years ago

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GREYUIT [131]

The answer is "b"
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2 years ago