You pick a card at random. Without putting the first card back, you pick a second card at random.

Fourteen of the 32 students in the class are boys. What is the ratio of boys to girls in the class?

AleksAgata [21]

The ratio would be "14:28"

8 0

3 years ago

What is the length of BC¯¯¯¯¯ ?

svlad2 [7]

Hello!

The length of side BC can be found by using the Law of Sines. This law states,

a / sin A = b / sin B = c / sin C.

Given:

Angle A = 51.2 degrees

Angle B = 60.3 degrees

Side AC (or B) = 21 cm

Side BC (or A) = ? cm

Let's plug in these values into the formula to find the missing side length.

a / sin (51.2) = 21 / sin (60.3)

a / sin (51.2) = 24.1759... (multiply both sides by sin(51.2)

a = 24.1759 · sin (51.2)

a = 18.841...

18.841 can be rounded to 19.

Therefore, the length of side BC is equal to 19 centimeters.

5 0

3 years ago

Line DE contains the points D (–1, –4) and E (3, 3). Line FG contains the point F (–3, 3). Which set of coordinates for point G

Bond [772]

Answer:

The set of coordinates for point G that make the two lines perpendicular are given through the relation

7y = -4x + 9

Step-by-step explanation:

The condition for perpendicularly of two lines is that the products of the slope of the two lines, m₁ and m₂ must be equal to -1.

For a line whose two points (x₁, y₁) and (x₂, y₂)

on the line are known, the slope of the line, m, is given as

m = (Δy/Δx) = (y₁ - y₂) ÷ (x₁ - x₂)

points D (–1, –4) and E (3, 3).

Let the slope of DE be m₁ and the slope of FG be m₂

Since the two lines are perpendicular,

m₁m₂ = -1

m₁ = (Δy/Δx) = (-4 - 3) ÷ (-1 -3) = (7/4) = 1.75

Since, m₁m₂ = -1, this means that,

m₂ = (-1/m₁) = (-1/1.75) = -(4/7)

Let the coordinates of point G be (x, y)

m₂ = (3 - y) ÷ (-3 - x)

(-4/7) = [(3-y)/(-3-x)]

-4(-3-x) = 7(3-y)

12 + 4x = 21 - 7y

7y = -4x + 9

Hope this Helps!!!

8 0

3 years ago

A rectangle is constructed with its base on theâ x-axis and two of its vertices on the parabola yequals=2525minusâxsquared2. wha

Nina [5.8K]

You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3

6 0

3 years ago

The area of a square boxing ring is 484 (feet squared). What is the perimeter of the boxing ring?

VARVARA [1.3K]

88 feet

<span>√</span>484=22
Perimeter is l+l+l+l
22+22+22+22=88

7 0

3 years ago

Read 2 more answers