How would you change 125.25% to decimal form?
2 answers:
You might be interested in
The answer is 1 + √2
sqrt(2) is
sqrt(3 + 2 sqrt 2) is
Now, let's simplify.
We will use square of sum: (a + b)² = a² + 2ab + b²
sqrt(2) is
sqrt(3 + 2 sqrt 2) is
Now, let's simplify
We will use square of sum: (a + b)² = a² + 2ab + b²
Answer:
See attached photo for graph
Step-by-step explanation:
Solve the equation for y... (that means get y by itself)
4y = -5x - 6 (subtract 5x from both sides)
y = -(5/4)x - 6/4 (divide both sides by 4)
y = -(5/4)x - 3/2 (simplify)
Now you can graph using slope-intercept method. The y-intercept is -3/2, and the slope is -5/4
Answer:
(a) 3.75
(b) 2.00083
(c) 0.4898
Step-by-step explanation:
It is provided that X has a continuous uniform distribution over the interval [1.3, 6.2].
(a)
Compute the mean of X as follows:
(b)
Compute the variance of X as follows:
(c)
Compute the value of P(X < 3.7) as follows:
Thus, the value of P(X < 3.7) is 0.4898.
<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.
1. Draw the angle bisector of angle JKL. The straight line in the green one.
2. Draw the angle bisector of angle KLJ. The straight line in the blue one.
3. Draw point D at these two lines.
4. Draw a line through point D perpendicular to side JK. The straight line in the red one.
5. Mark point G at the red line and side JK.
6. Draw a circle with center at D (point of concurrency) and a radius of DG.
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To circumscribe a circle in the attached triangle, we must follow the following steps:
1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.
2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.
3. The center of the circle is the intersection of these two lines
4. Draw a circle with center at D (point of concurrency).
<h3>3. Construct the two lines tangents to circle</h3>A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.