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3 years ago

Triangles ABD and ACE are similar right triangles. which ratio best explains why the slope of AB is the same as the slope of AC

Mathematics

2 answers:

sertanlavr [38]3 years ago

4 0

B. AC/EA = AB/DA because theyre equal slopes just different sizes

DochEvi [55]3 years ago

3 0

Answer:

The correct option is d.

Step-by-step explanation:

The slope of a line is

$Slope=\frac{Rise}{Run}$

Slope of line AB is

$m_{AB}=\frac{BD}{DA}$

Slope of line AC is

$m_{AC}=\frac{CE}{EA}$

Triangles ABD and ACE are similar right triangles, therefore the corresponding sides are proportional.

$\frac{AB}{AC}=\frac{BD}{CE}=\frac{AD}{AE}$

Using last two ratios, we get

$\frac{BD}{CE}=\frac{AD}{AE}$

The required ratio is

$\frac{BD}{DA}=\frac{CE}{EA}$

Therefore the correct option is d.

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Zucchini weights are approximately normally distributed with mean 0.8 pound and standard deviation 0.25 pound. Which of the foll

Charra [1.4K]

Answer:

0.81859

Step-by-step explanation:

We solve using z score method

The formula for calculating a z-score is is z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Mean 0.8 pound

Standard deviation 0.25 pound

For x = 0.55 pound

z = 0.55 - 0.8/0.25

= -1

Probability -value from Z-Table:

P(x = 0.55) = 0.15866

For x = 1.3 pounds

Z = 1.3 = 0.8/0.25

= 2

Probability value from Z-Table:

P(x = 1.3) = 0.97725

The probability that a randomly selected zucchini will weigh between 0.55 pound and 1.3 pounds is

P(x = 1.3) - P(x = 0.8)

0.97725 - 0.15866

0.81859.

6 0

3 years ago

Geometry help. Inscribing circle help and constructing lines.

prohojiy [21]

<h3>1. Inscribe a circle ΔJKL. 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.