Answer:
Sum of their present ages = 20 years
Step-by-step explanation:
Present age :
Neha's age = x years
Reemas' age = x - 4
After 5 years:
Neha's age = (x +5) years
Reemas' age = (x - 4 + 5 ) = x + 1
Sum of the ages of Reema and Neha will be 30
Therefore,
x + 5 + x + 1 = 30
x + x + 5 + 1 = 30
2x + 6 = 30
2x = 30 - 6
2x = 24
x = 24/2
x = 12
Neha's age = 12 years
Reema's age = 12 - 4 = 8 years
Sum of their present ages = 12 + 8 = 20 years
If my above assumption is correct, we can easily find the area of the left side of the rectangle by finding the area of 3 things. The first is the area of the entire region.
At = (0.8)(1.0+1.4) = 1.92
Next we need the area of the circle. Which is pi*r^2, with r being height divided by two. (0.4)
C = 0.502
Finally we need to find the area of the rectangle left region. Which is basically the rectangle from the center of the circle region to the end of the left of the entire region. This is just (1.0)(0.8) = 0.8
Finally we can find the area of the light blue area by just subtracting half the area of the circle (dark blue) from the area we just calculated.
0.8 - (0.502 / 2) = 0.549
Now for the probability we do this area divided by total area.
0.549/1.92 = 0.285 or about 0.29 which is just 29%
Answer:
see explanation
Step-by-step explanation:
Given a parabola in standard form
f(x) = ax² + bx + c ( a ≠ 0 )
Then the discriminant Δ = b² - 4ac informs us about the nature of the zeros
• If b² - 4ac > 0 then 2 real and irrational zeros
• If b² - 4ac > 0 and a perfect square then 2 real and rational zeros
• If b² - 4ac = 0 then 2 real and equal zeros
• b² - 4ac < 0 then zeros are not real
Given
f(x) = 3(x - 4)² - 12 ← expand factor using FOIL
= 3(x² - 8x + 16) - 12 ← distribute parenthesis by 3
= 3x² - 24x + 48 - 12
= 3x² - 24x + 36 ← in standard form
with a = 3, b = - 24 , c = 36 , then
b² - 4ac
= (- 24)² - (4 × 3 × 36
= 576 - 432
= 144 ← a perfect square
Then 2 zeros are real and rational and produce 2 x- intercepts
The two similar triangles are triangle BXY and triangle BAC. Since line XY is parallel to segment AC, angle BXY is congruent to angle A, and angle BYX is congruent to angle C. Since two corresponding angles are congruent, we use one of our similarity conjectures, Angle-Angle, to show these triangles are congruent.
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Hope this helps!
