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77julia77 [94]
3 years ago
11

Expand and simplify

0" id="TexFormula1" title="( \sqrt{3} + \sqrt{5} ) ^{2} " alt="( \sqrt{3} + \sqrt{5} ) ^{2} " align="absmiddle" class="latex-formula">

Mathematics
1 answer:
umka21 [38]3 years ago
7 0
Hey there !

Check the attachment.
Hope it helps you :)

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3⁷×3-⁴ I need help asap!​
SCORPION-xisa [38]

Hope it helps

Have a good day Ahead

8 0
3 years ago
Read 2 more answers
Julie is constructing a scale model of her room.The rectangular room is 10 1/4 inches by 8 inches.If 1 inch represents 2 feet of
zlopas [31]
The way to determine this is to know that 1 foot is 12 inches (so 2 is 24) 
Now the ration that would determine the scale factor of the room is 1:24 (1 inch for every 24 inches)
So the scale factor is 1:24
Now to determine the area we multiply the numbers we have by 2 and change the inches to feet (I hope that makes sense to you, it does to me, I'll show you)
10.25 * 2 = 20.5 ft.
8 * 2 = 16 ft.
now we know the dimensions of the room so we need to find the area.
A=B*H
20.5 * 16 = 328
so the area of the room is 328 ft.²
6 0
4 years ago
Read 2 more answers
Linda and Juan went shopping. Linda spent $14 less than Juan.
Rudiy27

Answer: Let x = amount Linda spent.

x=y-$18

Step-by-step explanation:

7 0
3 years ago
30 points please..I will report your answer if not right Thanks​
ella [17]
<h3>Answer:  SAS</h3>

=================================

How to get this answer:

We're told that AD = BC, so that is one pair of sides that are congruent. This forms the first "S" in "SAS"

The "A" refers to the congruent angles, which happen to be angle DAB and angle CBA, both are 90 degrees

The second "S" in "SAS" is the second pair of congruent sides. Those two sides are the overlapping shared side of AB. It might help to peel the two triangles apart to get a better look.

Note how the angles are between the two pairs of sides mentioned.

4 0
3 years ago
Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

rerange
\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
3 years ago
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