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olga2289 [7]
3 years ago
13

Three consecutive even numbers have a sum where one half of that sum is between 90 and 105.

Mathematics
1 answer:
Darya [45]3 years ago
5 0
n;\ n+2;\ n+4-three\ consecutive\ even\ numbers\\\\90 \ \textless \  \dfrac{1}{2}(n+n+2+n+4) \ \textless \  105\\\\90 \ \textless \  \dfrac{1}{2}(3n+6) \ \textless \  105\ \ \ |multiply\ both\ sides\ by\ 2\\\\180 \ \textless \  3n+6 \ \textless \  210\ \ \ |subtract\ 6\ from\ both\ sides\\\\174 \ \textless \  3n \ \textless \  204\ \ \ |divide\ both\ sides\ by\ 3\\\\58 \ \textless \  n \ \textless \  68\\\\Answer:\\60;\ 62;\ 64\ or\ 62;\ 64;\ 66;\ or\ 64;\ 66;\ 68\ or\ 66;\ 68;\ 70.
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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
8x-29 + 6x-11 + 3x-1 solve for x
elena-s [515]

Answer:

Step-by-step explanation:

I need help on this too.

5 0
3 years ago
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Distance between (-4,0) and (2,2)
svetlana [45]
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5 0
2 years ago
Elisa is making strawberry shortcake. One serving of strawberry shortcake requires one third cup of strawberries and one shortbr
Kisachek [45]

Given, To make a strawberry shortcake for one person, we need :

▪︎\frac{1}{3} cup of strawberry

▪︎one shortbread biscuit

Quantity of strawberries Elisa has = 3 cups

Number of shortbread biscuits she has = 12

Number of shortcakes she can make :

= \tt3 \div \frac{1}{3}

=  \tt\frac{3 \times 3}{1}

\color{plum} =\tt 9 \: people

Thus, Elisa can serve strawberry shortcakes to 9 people.

With 12 dozen shortbread biscuits she can serve 12 people.

Quantity of strawberries needed to serve 12 people :

=\tt \frac{1}{3}  \times 12

=\tt  \frac{12}{3}

\color{plum} = \tt4 \: cups

<h3>○=> Therefore :</h3>

▪︎Elisa can serve strawberry shortcakes to <u>9 people</u>.

▪︎Elisa will need <u>4 cups of strawberries</u> if she wants to use all dozen shortbread biscuits.

7 0
3 years ago
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