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Tanzania [10]
3 years ago
5

PLS how to do this 3 sums

Mathematics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

Step-by-step explanation:

7) PQ // RS,  RQ is transversal

x = 38 {Alternate interior angles are congruent}

In ΔPQR ,

x + y + ∠P  = 180   {Angle sum property of triangle}

38 + 90 + y = 180

  128 + y =  180

           y = 180 - 128

y = 52

9) Construct XY // AB

XY // AB  & AC is transversal

∠1 = ∠4     ---------(I)      {alternate interior angles are equal}

XY // AB  & BC is transversal

∠3 = ∠5     ---------(II)      {alternate interior angles are equal}

∠1 + ∠2 + ∠3 = 180   {straight line angles}

∠4 + ∠2 + ∠5 = 180    {From (I) & (II) }

Sum of three angles of triangle is 180.Hence proved.

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HELPPPP PLEASE HELP ME WITH BASIC MATH I REALLY APPRECIATE IT ((WILL MARK AS BRAINLIEST))​
Yakvenalex [24]

Answer:

3) 522

4) 468

Step-by-step explanation:

hello!

3) in the question, it says the cross section is a square, and the area of PQRS is 81.

for the square, we know that the length and height have to be the same, so if the area is 81, then we know the sides are both 9 because 9*9 is 81 and they're the same numbers multiplied. (same height and length)

with the information that all 4 sides of the square is 9 and the other side is labeled 10, we can figure out the rest.

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both 10 by 9 sides (top and bottom): 90+90=180

total surface area=162+180+180=522

4) for this problem, we can see that the triangle is a right triangle.

to solve the area of the triangle, we can do (9*12)/2=108/2=54.

there are two sides with triangles, so we double 54 (54*2=108)

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10 by 12 side: 120

now for the last slide, theres no information on the length. the formula for the diagonal is

a^2+b^2=c^2

which is

9^2+12^2=81+144=255=15^2

with this, the area of the rectangular side gives us 10*15 which is 150.

adding it together, we get

108+90+120+150=468.

hope this helps!

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3 years ago
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Answer:

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3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\begin{pmatrix} \frac{3}{3}  & \frac{0}{3}  & \frac{-3}{3} \\\\ \frac{6}{3}  & \frac{3}{3}  & \frac{0}{3} \\\\ \frac{9}{3}  & \frac{6}{3}  & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1  & 0 & - 1\\ 2  & 1 & 0 \\ 3  & 2  & 1 \end{pmatrix}}\\

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3 years ago
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