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ValentinkaMS [17]
3 years ago
15

How many rainbow six events has there been since beta?

Mathematics
1 answer:
Pie3 years ago
8 0
There has been 25 I believe.
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A circle has radius of 18 cm. Find the area of the smaller of the two regions determined by a chord with length of 18√2 cm . Hin
boyakko [2]
The area is 81*pi-162. Look at the triangle formed by the chord and the two radii connected to the endpoints of the chord. Since 18^2+18^2=(18*\sqrt2)^2, this is an isosceles right triangle by Pythagorean theorem. The area of the quarter circle containing the region we look for is pi*18^2/4=81*pi. The area of the right triangle is 18*18/2=162. The difference between these two areas, or 81*pi-162, is the area of the smaller region.
8 0
3 years ago
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in a 7th grade class. There are 4 boys for every 5 girls, How many girls are in the class if there are 20 boys in the class
spin [16.1K]

Answer:

<u>Direct proportion</u>

4 boys = 5 girls

20 boys = 25 girls

4 0
3 years ago
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BRAINLIEST PLUS POINTS
vovangra [49]

Answer:

a) e(0,2)

b) 1

c) read below

d) read below

Step-by-step explanation:

Midpoint:

to calculate the midpoint e of a segment

e((x1+x2)/2, (y1+y2)/2)

so (-1+1)/2 and (3+1)/2  

e(0,2)

Slope formula:

the slope of a straight line between two points is

\frac{y_2-y_1}{x_2-x_1}

so between e and B it's equal to (4-2)/(2-0) = 1

Condition for perpendicularity : slope(AC) = -1/slope(eB) = -1

so we calculate the slope of a straight line through A and C

(1-3)/1-(-1) = -2/2 = -1   so they are perpendicular

Area:

eB is the height of the triangle as it's perpendicular to the base, so applying the standard formula Area = (AC*eB)/2 we can find the area

6 0
3 years ago
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3 years ago a father was 4 times as old as his daughter is now. The product of their present ages is 430. calculate the present
iren [92.7K]

If the product of the present age of father and his daughter is 430 and the age of father was 4 times as old as his daughter is now, then the present age of father is 43 and the present age of daughter is 10.

Given that the product of the present age of father and his daughter is 430 and the age of father was 4 times as old as his daughter is now.

We are required to find the present age of the father and daughter.

Suppose the present age of the father is x.

Suppose the present age of his daughter is y.

We are given product of present ages be 430.

xy=430----------1

y=430/x

According to question,

x-3=4y

Put the value of y=430/x.

x-3=4*430/x

x^{2}-3x=1720

x^{2} -3x-1720=0

x=(3±\sqrt{3^{2}-4*1*(-1720) })/2*1

x=3±\sqrt{6889})/2

x=(3±83)/2

x=(3+83)/2

(Ignoring the negative value because the age cannot be negative as it is coming before the greater number)

x=43

y=430/x

y=430/43

y=10

Hence if the product of the present age of father and his daughter is 430 and the age of father was 4 times as old as his daughter is now, then the present age of father is 43 and the present age of daughter is 10.

Learn more about product at brainly.com/question/10873737

#SPJ9

5 0
2 years ago
ΔABC with vertices A(-3, 0), B(-2, 3), C(-1, 1) is rotated 180° clockwise about the origin. It is then reflected across the line
Nonamiya [84]
Okay, so im not 100% sure on this, but, A(0,-3) B(3,-2) C(1,-1), again, not 100%, but only because of the reflection across the y=-x
4 0
4 years ago
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