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2 years ago

In the diagram below, BC is an altitude of ABD. To the nearest whole unit, what is the length of CD?

Mathematics

2 answers:

dem82 [27]2 years ago

7 0

First we need to calculate the 2 angles that make up Angle B at the top of the triangle.
Using that angle we can find the length of CD.

See the attached picture:

Lina20 [59]2 years ago

5 0

Answer:

D. 26.

Step-by-step explanation:

We have been given an image of a triangle. We are asked to find the length of CD.

We will use altitude on hypotenuse theorem to solve our given problem.

$\frac{CD}{\text{Altitude}}=\frac{\text{Altitude}}{AC}$

Upon substituting our given values, we will get:

$\frac{CD}{37}=\frac{37}{53}$

$\frac{CD}{37}\times 37=\frac{37}{53}\times 37$

$CD=\frac{37\times 37}{53}$

$CD=\frac{1369}{53}$

$CD=25.830188679\approx 26$

Therefore, the length of CD is 26 units and option D is the correct choice.

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2 years ago

Ahmed doubled his savings, then deposited another hundred dollars. Let s represent the amount he originally had in savings. Whic

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Do the opposite of what you did in reverse order.

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2 years ago

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Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir

lys-0071 [83]

Answer:

a) $\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

b) $Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

c) $\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

If we apply this formula for the four methods we got:

$\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

$Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

$\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

3 0

1 year ago

For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

natulia [17]

<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let $e$ the cost of the exam hours

Let $w$ be the cost of the workshop hours

Let $l$ be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

$e=2w$ equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

$w=2l$ equation (2)

Finally, we also know that 3hr exams 24hr workshops and 12hr lectures cost $528, so:

$3e+24w+12l=528$ equation (1)

Now, lets find the value of $l$ :

Step 1. Solve for $l$ in equation (3)

$3e+24w+12l=528$

$12l=528-3e-24w$ equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

$12l=528-3e-24w$

$12l=528-3(2w)-24w$

$12l=528-6w-24w$

$12l=528-30w$ equation (5)

Step 3. Replace equation (2) in equation (5) and solve for $l$

$12l=528-30w$

$12l=528-30(2l)$

$12l=528-60l$

$72l=528$

$l=\frac{528}{72}$

$l=\frac{22}{3}$

$l=7.33$

Cost of lectures = $7.33 per hour

3 0

2 years ago

Read 2 more answers

Approximately what is the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertica

Lera25 [3.4K]

Answer:

212m

Step-by-step explanation:

The set up will be equivalent to a right angled triangle where the height is the opposite side facing the 45° angle directly. The length of the rope will be the slant side which is the hypotenuse.

Using the SOH, CAH, TOA trigonometry identity to solve for the length of the rope;

Since we have the angle theta = 45° and opposite = 150m

According to SOH;

Sin theta = opposite/hypotenuse.

Sin45° = 150/hyp

hyp = 150/sin45°

hyp = 150/(1/√2)

hyp = 150×√2

hyp = 150√2 m

hyp = 212.13m

Hence the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertical height of 150 m is approximately 212m

8 0

2 years ago