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Volgvan
2 years ago
14

QUESTION 5 PLS HELPPP MEE

Mathematics
1 answer:
Anton [14]2 years ago
8 0

Answer:

C

Step-by-step explanation:

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Help with this problem please!!!!!
mariarad [96]

Answer:

3

Step-by-step explanation:

3 0
3 years ago
Solve for x<br>2x²-3x-209=0​
kifflom [539]

Answer:

x= 11

Step-by-step explanation:

3 0
3 years ago
Help anyone can help me do the question 4,I will mark brainlest.​
nasty-shy [4]

Answer:

Hello,

Step-by-step explanation:

A=(1,2)\\B=(0,-1)\\\overrightarrow{AB}=((0,-1)-(1,2)=(-1,-3)\ ||\overrightarrow{AB}||^2=1+9=10\\\overrightarrow{BC}=((3,-2)-(0,-1)=(3,-1)\ ||\overrightarrow{BC}||^2=9+1=10\\\\Triangle\ is\ isosceles.\\\\\overrightarrow{AB}.\overrightarrow{BC}=(-1,-3)*\left[\begin{array}{c}3\\-1\end{array}\right] =-3+3=0\\\\Triangle \ is\ right.\\\\

8 0
2 years ago
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Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
The rate of change in the function y=x+4 is________ the rate of change of the function represented in the table.
Snezhnost [94]
Recall that the slope is the "average rate of change" of any function.

\bf y=\stackrel{slope}{1}x+4\qquad \qquad &#10;\begin{array}{ccll}&#10;x&y\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;0&6\\2&8\\4&10\\6&12&#10;\end{array}&#10;\\\\\\&#10;\textit{let's check what is the slope of the tabled function}\\&#10;\textit{by using two points from it}

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;&({{ 2}}\quad ,&{{ 8}})\quad &#10;%   (c,d)&#10;&({{ 6}}\quad ,&{{ 12}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{12-8}{6-2}\implies \cfrac{4}{4}\implies 1
4 0
3 years ago
Read 2 more answers
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