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vivado [14]
2 years ago
11

Two birds sit at the top of two different trees. The distance between the first bird and a birdwatcher on the ground is 29.7 fee

t. The distance between the birdwatcher and the second bird is 42.1 feet.
What is the angle measure, or angle of depression, between this bird and the birdwatcher?

Two trees that are 29 point 7 feet tall with a bird on top of each. Distance between birdwatcher and first bird is 42 point 1 feet. Angle of depression between birdwatcher and first bird is unknown.

45.1°
44.9°
54.8°
35.2°
Mathematics
1 answer:
icang [17]2 years ago
8 0

Answer: 44.9

Step-by-step explanation:

flvs

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Find the midpoint of the segment with the given endpoints.<br> (-4,-5) and (10,- 9)
Elanso [62]

Answer:

(3, -7)

Step-by-step explanation:

\left(\frac{-4+10}{2}, \frac{-5-9}{2} \right)=(3, -7)

6 0
1 year ago
Solve for x d(-3+x)= kx+9
olga nikolaevna [1]
1. Write the equation:
d(-3 + x) = kx + 9
2. "Open" the parenthesizes:
-3d + dx = kx + 9
3. Take kx to the left side, -3d to the right, everything with different sign (+ replace with - when transferring, - replace with +):
dx - kx = 9 + 3d
4. Factor x in the left side:
x(d - k) = 9 + 3d
5. Finally, divide the whole equation by (d - k):
x = (9 + 3d)/(d - k)
That's the final answer. Good luck!
4 0
3 years ago
Read 2 more answers
If ΔABC ≅ ΔDEF, then what corresponding parts are congruent?
GaryK [48]

The corresponding parts that are congruent are (a) AB and DE

<h3>How to determine the congruent parts?</h3>

The statement ΔABC ≅ ΔDEF means that the triangles ABC and DEF are congruent.

This implies that the following points are corresponding points:

A and D; B and E; C and F

When two corresponding points are joined together, the congruent parts are:

AB and DE, AC and DF, BC and EF

Hence, the corresponding parts that are congruent are (a) AB and DE

Read more about congruent triangles at:

brainly.com/question/1675117

#SPJ1

3 0
2 years ago
What's the area?<br><br> C= pi/2 <br><br> A=____
kompoz [17]
The answer is actually 1.571
6 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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