The point (2, -5) would not be one of the correct vertices of the reflected shape.
When we reflected a shape across the line y=x, we just replace the x's with y's and y's with x's.
Our new points are (5, 2) (4, 6) and (3, 3). None of the points contain negative numbers.
Answer:
<h3>X<5 and X<3</h3>
Step-by-step explanation:
To solve this problem, first, you have to isolate it on one side of the equation. Remember to solve this problem, isolate x on one side of the equation.
x+3<8 and 3(x+4)-11<10
x+3<8
x+3-3<8-3 (First, subtract 3 from both sides.)
8-3 (Solve.)
8-3=5
x<5
3(x+4)-11<10
3(x+4)-11+11<10+11 (Add 11 from both sides.)
10+11 (Solve.)
10+11=21
3(x+4)<21
3(x+4)/3<21/3 (Next, divide by 3 from both sides.)
21/3 (Solve.)
21/3=7
x+4<7
x+4-4<7-4 (Then, subtract 4 from both sides.)
7-4=3
x<3
The correct answer is x<5 and x<3.
Answer:
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In Indian Server it is common.
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%
