12) Why doesn't SSA prove two triangles are congruent? *

1 answer:

7 0

Ans: Four shortcuts allow students to know two triangles must be congruent: SSS, SAS, ASA, and AAS. Knowing only side-side-angle (SSA) does not work because the unknown side could be located in two different places. The same is true for side angle side, angle side angle and angle angle side.

You might be interested in

Combine the like terms of the expression (8 + 6x) + (-4 + 9c). Explain how you simplified the expression in each step. * pls hel

Delicious77 [7]

Answer: 6x+9c+4

Step-by-step explanation: (8 + 6x) + (-4 + 9c)

8+6x-4+9c

6x+9c+8-4

6x+9c+4

8 0

3 years ago

Graph the following piecewise function on a separate piece of graph paper and upload your graph below.

Alecsey [184]

ANSWER

To graph the function

$f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x$

follow the steps below.

1. Find y- intercept by plugging in $x=0$ .

$x=0$ is on the interval, $-4\leq x$ , so we substitute in to

$f(x)=x+4$

$\Rightarrow f(0)=0+4$

$\Rightarrow f(0)=4$

Hence the y-intercept is $(0,4)$

2. Find x-intercept by setting $f(x)=0$

This implies that

$x+4=0,$ on $-4\leq x$

or

$2x-1=0$ on $3\leq x$

We now solve for $x$ on each interval,

$x=-4,$ on $-4\leq x$

or

$x=\frac{1}{2}$ on $3\leq x$

But observe that

$x=\frac{1}{2}$ does not belong to $3\leq x$

This means it can never be an intercept for this piece-wise function.

Hence our x-intercept is $(-4,0)$

3. Plotting the boundaries of the interval.

For $f(x)=x+4$ on $-4\leq x$

$f(-4)=-4+4$

$\Rightarrow f(-4)=0$ .

This point $(-4,0)$ coincides with the x-intercept.

$f(3)=3+4$

$f(3)=7$

So we have the point $(3,7)$ . But note that $x=3$ does not belong to this interval so we plot this point as a hole.

For $f(x)=2x-1$ on $3\leq x$

$f(3)=2(3)-1$

$\Rightarrow f(3)=5$

So we plot $(3,5)$

$f(6)=2(6)-1$

$\Rightarrow f(6)=11$

So we plot $(6,11)$ also as a hole.

Plotting all these points we can now graph the function,

$f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x$

See attachment for graph.