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Irina18 [472]
3 years ago
5

Quadrilateral ABCD has vertices at A(-2, -4), B(1, -1), C(-2, 2) and D(-5, -1). What is the name of the figure?

Mathematics
1 answer:
dimaraw [331]3 years ago
4 0

Answer:

a square

Step-by-step explanation:

bc if u plot them on a graph, you get a rhombus but when you turn the page, you also get a square. it said under the rhombus, not a square, so we cross that one out and we hv the res. all the sides are equal so the only answer left is a square

hope it helped xx

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A

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What is m&lt;15 ? <br> A. 80 <br> B. 70 <br> C. 100<br> D. 110
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7 0
2 years ago
Write the following equation in slope -intercept form : y - 3 = 4(x + 2)​
andreyandreev [35.5K]

Answer:

y=4x+11

Step-by-step explanation:

multiply 4 into the parentheses to get y-3=4x+8 then invert -3 so add three to both sides to get rid of it so instead of +8 its +11 so now its y=4x+11 and slope intercept is y=mx+b, so 4 is m, slope, and 11 is b, the y intercept i hope that helps

7 0
3 years ago
3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl
riadik2000 [5.3K]

Elimination Method

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}

If we multiply the equation 3 by (-1) we obtain this:

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\  \end{gathered}

2. We can replace this value of X in the 1st and 2nd equations

\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}

3. If we simplify:

\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}

4. We can obtain Y from this two equations:

\begin{gathered} Y=-\frac{-12Z+58}{9} \\  \end{gathered}

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}

6. If we simplify, we obtain:

X=\frac{-3Z+7}{9}

7. In conclusion, we obtain that

(X,Y,Z) =

(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)

8 0
1 year ago
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