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3 years ago

What is the quotation -4/5 divided by 2

Mathematics

2 answers:

rodikova [14]3 years ago

5 0

Answer:

-0.4

Step-by-step explanation:

vekshin13 years ago

4 0

Exact form : -2/5
Decimal form : -0.4

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PLEASE HELP!!! I’LL GIVE BRAINIEST!!!

Vilka [71]

Answer:

8x - 17 = 5x +19

Step-by-step explanation:

Corresponding angles are congruent

thus, 8x - 17 = 5x +19

Hope it helps!

6 0

2 years ago

Read 2 more answers

If f(x) = x³ + x, find <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%2C%20f%5Cneq%200" id="TexForm

Snowcat [4.5K]

I think you are doing limits so this is what I did
and that's how I factored using the box method because it's easier to track distribution.

6 0

3 years ago

Which equation shows y = (x − 4) 2 − 3 written in standard form?

Mrac [35]

Answer:

2x-y=11 is the standard form

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP ASAP !! IF I DONT FINISH THIS HOMEWORK ILL GET GROUNDED .

kvv77 [185]

Answer:

A) Histogram

Step-by-step explanation:

A) The first step would be to look at the purpose and use for each type of plotting method;

Dot Plot: Used to represent the distribution of data (for ex; #of Strawberries, Blueberries, and Raspberries.

Histogram: A histogram is used to summarize discrete or continuous data. In other words, it provides a visual interpretation of numerical data by showing the number of data points that fall within a specified range of values

Box Plot: Summerizes a set of data measured on an interval scale.

Best choice: Histogram- The reason why a histogram is the best representation of the student quartiles is because a histogram is used to summarize discrete or continuous data, and the given data is discrete

To create your histogram you first have to create a frequency table like the one below;

On the vertical axis, place frequencies. Label this axis "Frequency".

On the horizontal axis, place the lower value of each interval. Label this axis with the type of data shown (Score, etc.)

Draw a bar extending from the lower value of each interval to the lower value of the next interval. The height of each bar should be equal to the frequency of its corresponding interval.

That's how it's done!

5 0

3 years ago

Solve the following biquadratic equation: <img src="https://tex.z-dn.net/?f=%28%287%21%29%2Ax%5E4%29%2B%28%285%21%29%2Ax

jekas [21]

$\large\begin{array}{l} \textsf{Solve the equation}\\\\ \mathsf{7!\cdot x^4+5!\cdot x^2-3=0}\\\\ \mathsf{7!\cdot (x^2)^2+5!\cdot x^2-3=0}\\\\\\ \textsf{Substitute}\\\\ \mathsf{x^2=t\quad(t\ge 0)}\\\\\\ \textsf{so the equation becomes}\\\\ \mathsf{7!\cdot t^2+5!\cdot t-3=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=7!}\\\mathsf{b=5!}\\\mathsf{c=-3} \end{array} \right. \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(5!)^2-4\cdot 7!\cdot (-3)}\\\\ \mathsf{\Delta=(5!)^2+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7\cdot 6\cdot 5!}\\\\ \mathsf{\Delta=5!\cdot (5!+12\cdot 7\cdot 6)}\\\\ \mathsf{\Delta=5!\cdot (120+504)}\\\\ \mathsf{\Delta=5!\cdot 624} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=120\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3\cdot 5)\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^{3+4}\cdot 3\cdot 5 \cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^7\cdot 3^2\cdot 5 \cdot 13} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=2^{6+1}\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{6}\cdot 2\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{3\,\cdot\,2}\cdot 3^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 130} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(5!)\pm \sqrt{(2^3\cdot 3)^2\cdot 130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-120\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-2^3\cdot 3\cdot 5\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 4\cdot 7\cdot (2\cdot 3)\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 2^2\cdot 3\cdot 7\cdot 2\cdot 5\cdot 3!}} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{(2^3\cdot 3)\cdot \big(\!\!-5\pm \sqrt{130}\big)}{(2^3\cdot 3)\cdot 7\cdot 2\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{7\cdot 2\cdot 5\cdot 3\cdot 2\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 7\cdot 5\cdot 3\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}&~\textsf{ or }~& \end{array} \mathsf{t=\dfrac{-5-\sqrt{130}}{2^2\cdot 105}}\quad\textsf{(useless, since }\mathsf{t\ge 0}\textsf{)}\\\\ \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \textsf{Substituite back for }\mathsf{x^2=t:}\\\\ \mathsf{x^2=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}\\\\ \mathsf{x=\pm \sqrt{\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}}\\\\ \mathsf{x=\pm \dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}\\\\ \boxed{\begin{array}{rcl} \mathsf{x=-\,\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}&~\textsf{ or }~&\mathsf{x=\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Solution: }\mathsf{S=\left\{-\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}},\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}}\right\}.} \end{array}$

If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2094277

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$

Tags: solve biquadratic equation factorial Bhaskara solution

4 0

3 years ago