Answer:
Distance between two points= √(x2-x1)²+(y2-y1)²
D= 15units
Let "k" be the y-coordinate of A
A(-6,k)
B(3,2)
15= √(3--6)²+(2-k)²
Taking square of both sides to eliminate the square root.
15²=(9)²+(2-k)²
225=81+4-4k+k²
k²-4k - 140=0
Using Quadratic Formula to evaluate
k= 14 or k=-10
The Possible coordinates of A are
(-6,14) or (-6,-10)
Answer:
what grade is this lol I don't know
Answer:
\mathrm{Domain\:of\:}\:x^3+3x^2-x-3\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}
Step-by-step explanation:
Answer:
The area is 21 inches
Step-by-step explanation:
3x3=9
3x2=6
9+6=15 ( inches in the white box )
6x6=36 ( area of the whole box)
36-15= 21 ( inches in the shaded area )
You can use this formula <span>P(AorB) = P(A) + P(B) - P(AandB)
Given:
35 LG (14 F & 21 M)
44 SB (28 F & 16 M)
Req:
- the probability that it is a female (F) or a sky blue (SB)
Sol:
</span>P(F or SB) = P(F) + P(SB) - P(F and SB)
= [(14 F + 28 F)/(35 + 44)] + [(44 SB)/(35 + 44)] - [(28 F)/(35 + 44)]
= 53.16 + 55.70 - 35.44
= 73.42%
You have to deduct 28 female parakeets from 44 sky blue parakeets because the 28 parakeets are already accounted for in the female parakeets. You can also think of how many ways you can choose a female parakeet and a sky blue parakeet. Add all female parakeets (14 + 28) = 42. Sky blue parakeet equaled to 44. Minus the 28 female parakeets included in the sky blue parakeet to avoid double counting. 42 + 44 - 28 = 58 divided by 79 (35 + 44) total parakeets = 73.42%