Can a quadrilateral be a rhombus ?

1 answer:

6 0

Yes it can. It doesn't <em>have to be</em>, but it can be without too much trouble.

In fact, ALL rhombuses are quadrilaterals.

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If mce =40 and m cde =30 what is mab

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The measure of arc AB is 100°.

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Solve x2 + 8x − 3 = 0 using the completing-the-square method.

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$(a+b)^2=a^2+2ab+b^2\\------------------\\\\x^2+8x-3=0\ \ \ |add\ 3\ to\ both\ sides\\\\x^2+2x\cdot4=3\ \ \ |add\ 4^2\ to\ both\ sides\\\\x^2+2x\cdot4+4^2=3+4^2\\\\(x+4)^2=3+16\\\\(x+4)^2=19\iff x+4=-\sqrt{19}\ or\ x+4=\sqrt{19}\\|subtract\ 4\ from\ both\ sides\\\\\boxed{x=-4-\sqrt{19}\ or\ x=-4+\sqrt{19}}$

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Read 2 more answers

Find an equation for the line below. Thanks !

daser333 [38]

Answer:

$\huge\boxed{y=\frac{1}{4}x + \frac{13}{4}}$

Step-by-step explanation:

In order to find the equation to this line, we need to note that we see two points on this graph. Using these two points, we can use them to find the slope of the graph and use one to find the y-intercept.

We know that the slope of a line is defined by $\frac{\Delta y}{\Delta x}$ (change in y / change in x). Therefore, we can use our two points that we know - (-5, 2) and (3, 4) to find the slope.

The change in y is $4 - 2 = 2$ , and the change in x is $3 - (-5) = 3+5= 8$ . Therefore, our slope is $\frac{2}{8} = \frac{1}{4}$ .

Now that we know our slope, our equation in slope-intercept form looks something like this.

$y = \frac{1}{4}x + b$

However, we still have b to solve for. We can solve for this by substituting a point we already know into the equation. Let's substitute (3, 4) inside.

$4 = \frac{1}{4} \cdot 3+b$
$4 = \frac{3}{4} + b$
$b = \frac{16}{4}- \frac{3}{4}$
$b = \frac{13}{4}$

So now we know that the y-intercept is $\frac{13}{4}$ . Plugging that into our equation finishes it off, leaving our final equation to be $y = \frac{1}{4}x + \frac{13}{4}$ .