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bogdanovich [222]
3 years ago
9

????DAY is a rhombus. If point ???? has coordinates (2, 6) and ???????? has coordinates (8, 10), what is the

Mathematics
1 answer:
satela [25.4K]3 years ago
7 0

Please dont put question mark on these they dont make sense so make thi sbetter!

Please!

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Solve for y please help!!!!!
mrs_skeptik [129]

Answer:

148

Step-by-step explanation:

3p + 85 +2p -10 =180

5p+75=180

5p=105

p=21

3p+85

3(21)+85=148

3 0
3 years ago
Solve the following matrix equations: (matrices)
Masja [62]

Step-by-step explanation:

a)

3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} \\  \\  3X  = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} -  \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}  \\  \\ 3X  = \begin{pmatrix}  - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\  \\ 3X  = \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \frac{1}{3} \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \begin{pmatrix}  \frac{ - 3}{3}  & \frac{3}{3}  \\  \\ \frac{6}{3}  & \frac{9}{3} \end{pmatrix}\\  \\ \huge \red{ X}  =  \purple{ \begin{pmatrix}  - 1  &1  \\ 2 & 3 \end{pmatrix}}

b)

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}}\\

8 0
3 years ago
340 pages; 20% decrease=? (With work plz)
pishuonlain [190]
340 × .2 = 68

68 page decrease

340 - 68 = 272

272 pages left
4 0
3 years ago
A candy maker is making candy canes to sell over the holidays she has 45 left over the last month sale and she can make 225 per
notsponge [240]

F(x) stands for a function, and if x stands for the number of days, than she can do 225 per day, which would be the multiplication to find the toatl 225x.

Then, with 45 extra, we add that to the equation:

F(x) = 225x + 45

Answer: F(x) = 225x + 45

Credit to: @2Educ8

+ = <3

8 0
3 years ago
A student determined that the area of the segment of c shown above is Asegment = 137.71 ft2. The student's work is shown below.
ivolga24 [154]

Answer:

Option D. The student did not use the correct formula to calculate the area of the segment

Step-by-step explanation:

step 1

Find the area of the isosceles triangle

Applying the law of sines

A=\frac{1}{2}(12^{2})sin(60\°)=62.35\ ft^{2}

step 2

Find the area of the sector

The area of the sector is 1/6 of the area of the circle

so

A=\pi r^{2}/6

substitute the value

A=(3.14)(12)^{2}/6=75.36\ ft^{2}

step 3

Find the area of the segment

The area of the segment is equal to the area of sector minus the area of triangle

A=75.36\ ft^{2}-62.35\ ft^{2}=13.01\ ft^{2}

therefore

The student did not use the correct formula to calculate the area of the segment

4 0
3 years ago
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