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2 years ago

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]2 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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2. Consider a sequence where f(1) = 1,f(2) = 3, and

Darya [45]

Answer:

24,27,30 and 33 and so on

Step-by-step explanation:

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2 years ago

How do I solve 5x+1=7x+19

notsponge [240]

Answer:

x=-9

Step-by-step explanation:

5x+1=7x+19

1 = 7x-5x +19

1-19 = 2x

-18/2 = x

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Hope this helps!

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5 0

3 years ago

What is an equation for the line that passes through (-2,-3) with a slope of -1/2?

Crazy boy [7]

Answer:

Step-by-step explanation:

To write the equation of a line use the point slope form of a line substituting m = -1/2 and the point (-2,-3).

$y -y_1 = m(x-x_1)\\y --3=-\frac{1}{2}(x--2)\\y+3 = -\frac{1}{2}(x+2)$

Convert to standard form by applying the distributive property and rearranging the terms.

$y + 3 = -\frac{1}{2}(x+2)\\y+3 = -\frac{1}{2}x -1\\2y + 6 = -x -2\\x+2y + 6 = -2\\x + 2y = -8$

The equation x +2y = -8 is the same equation as 4y + 2x = -16 just doubled----> 2x + 4y = -16

8 0

2 years ago

To the nearest degree what is the measure of each exterior angle of a regular hexagon ?

otez555 [7]

This is the concept of geometry, the total sum of angles in a hexagon is given by:
(n-2)*180
=(6-2)*180
=720
the size of each interior angle will be:
120°
N/B
the exterior and interior angles are supplementary to each other.
Therefore the size of the exterior angle will be:
180-120
=60°
the answer is A

3 0

3 years ago

Read 2 more answers

What is the limit as x is approaching 0 of the absolute value of x

vesna_86 [32]

Recall that, an absolute value expression is a piece-wise in effect, thus we'd check the single-sided limits,

$\bf \lim\limits_{x\to 0}~|x|\implies 
\begin{cases}
\lim\limits_{x\to 0^-}~-x\\\\
\lim\limits_{x\to 0^+}~+x
\end{cases}\implies 
\begin{cases}
0\\\\
0
\end{cases}\implies \lim\limits_{x\to 0}~|x|=0$

6 0

3 years ago