Given that ΔBDA is similar to ΔBDC and:
AD≡DC
AB≡BC
BD≡BD (shared side)
then the best postulate to use is the side-side-side (SSS) postulate.
Answer: SSS
Answer: The perimeter is 48 inches. The area is 140
Step-by-step explanation:
Since we have the length and width, we know the formula of perimeter is L+L+W+W. So 10+10+14+14= 48 The formula of area is LxW so 14 x 10= 140
Answer:
f(-3)= 19
Step-by-step explanation:
you have to use substitution. Every time that you see an x, plug it in for a -3
f(x)= -5x+4
f(-3)= -5(-3)+4
f(-3)= 19
Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.
Because we know 3 and -1 (multiplicity 2) are both roots, the last root
is such that we can write

There are a few ways we can go about finding
, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be
.
Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

so the missing root is 3.
Other things we could have tried that spring to mind:
- three rounds of division, dividing the quartic polynomial by
, then by
twice, and noting that the remainder upon each division should be 0
- rational root theorem
Answer: (-1, 2)
Step-by-step explanation:
<u>It's a counter-clockwise rotation, that means (x, y) changes to (y, -x).</u>
(-2, -1) ⇒ (-1, -(-2)) ⇒ (-1, 2)
<u>If it's a clockwise rotation, then (x, y) will change to (-y, x)</u>
(-2, -1) ⇒ (-(-1), -2) ⇒ (1, -2)