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Stella
14 days ago
12

George is 1.94 meters tall and wants to find the height of a tree in his yard. He started at the base of the tree and walked 10.

20 meters along the shadow of the tree until his head was in a position where the tip of his shadow exactly overlaps the end fo the tree top's shadow. He is now 5.1 meters from the end of the shadows. How tall is the tree?

Mathematics
1 answer:
Leona [9.2K]14 days ago
7 0
Create a visual representation to depict the situation outlined below.

Let h signify the height of the tree.

Since ΔABC is similar to ΔADE, it follows that
DE/BC = AD/AB

This means that
h/1.94 = (5.1 + 10.2)/5.1 = 3
Consequently, h = 1.94*3 = 5.82 m

Final result: 5.82 m

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Using the extended Euclidean algorithm, find the multiplicative inverse of a. 1234 mod 4321 b. 24140 mod 40902
AnnZ [9099]

(a) The multiplicative inverse of 1234 (mod 4321) is x so that 1234*x ≡ 1 (mod 4321). We can apply Euclid's algorithm:

4321 = 1234 * 3 + 619

1234 = 619 * 1 + 615

619 = 615 * 1 + 4

615 = 4 * 153 + 3

4 = 3 * 1 + 1

Now we will express 1 as a linear combination of 4321 and 1234:

1 = 4 - 3

1 = 4 - (615 - 4 * 153) = 4 * 154 - 615

1 = 619 * 154 - 155 * (1234 - 619) = 619 * 309 - 155 * 1234

1 = (4321 - 1234 * 3) * 309 - 155 * 1234 = 4321 * 309 - 1082 * 1234

This reduces to

1 ≡ -1082 * 1234 (mod 4321)

Thus, the inverse is

-1082 ≡ 3239 (mod 4321)

(b) Since both 24140 and 40902 are even, their GCD cannot equal 1, indicating no inverse exists.

8 0
1 month ago
What is the domain of the function graphed below?
zzz [9080]
The function is applicable within the segments of x:
(-∞, -1) and [-1, 7), meaning it is valid for x < 7.

Importantly,
the function cannot be evaluated at x = -1 in the left part of the linear graph, while it is valid at x = -1 in the right segment of the same line. Additionally, the function is not defined at x = 7 or any value above it.

Conclusion: x < 7.
5 0
1 month ago
Read 2 more answers
The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the fun
zzz [9080]

Average fall rate =

\frac{h(3)-h(0)}{3}Detailed explanation:

Given that the position of the object is expressed as a function of time, we can compute the average velocity of the object during the first 3 seconds by determining its position at time 0 and at 3 seconds, finding the displacement over that period, and then dividing this distance by the elapsed time (3 seconds). This approach follows the concept that velocity equals distance covered divided by time spent:

Average rate of fall = (h(3) - h(0))/ 3

Average rate of fall =

4 0
10 days ago
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A truck is carrying 10 3 bushels of apples, 10 2 bushels of grapes, and 10 1 bushels of oranges. Each bushel of apples weighs
Svet_ta [9500]

Answer:

10,088 pounds

Step-by-step explanation:

Provided data

103 bushels of apples

102 bushels of grapes

101 bushels of oranges

With the respective weight per bushel:

Apples = 32 pounds

Grapes = 25 pounds

Oranges = 42 pounds

The total weight can be calculated by summing the products of each type:

Total Weight: 103(32) + 102(25) + 101(42) = 10,088 pounds

5 0
8 days ago
W hich of these scales is equivalent to the scale 1 cm to 5 km? Select all that apply.  (Lesson 1-11) A. 3 cm to 15 km D.    5 m
AnnZ [9099]

The corresponding scales are:

3 cm for 15 km translates to 1 cm for 5 km ⇒ answer A

5 mm for 2.5 km translates to 1 cm for 5 km ⇒ answer D

1 mm for 500 m translates to 1 cm for 5 km ⇒ answer E

Detailed explanation:

To engage with this problem:

1. Given that 1 cm is equivalent to 5 km, all provided answers must

 be converted to cm and km

2. Determine the matching scales against the provided scale

A.

Since 3 cm indicates 15 km

- Divide both values by 3

Thus, 1 cm indicates 5 km

3 cm for 15 km equates to 1 cm for 5 km

B.

Since 1 mm signifies 150 km

- Convert mm to cm

Since 1 cm = 10 mm

Thus, 1 mm = \frac{1}{10} cm = 0.1 cm

So, 0.1 cm indicates 150 km

- Then multiply both by 10

So, 1 cm indicates 1500 km

1 mm to 150 km does not match with 1 cm to 5 km

C.

Since 5 cm signifies 1 km

- Divide both by 5

Thus, 1 cm signifies 0.2 km

5 cm to 1 km does not match with 1 cm to 5 km

D.

Since 5 mm signifies 2.5 km

- Convert mm to cmSince 1 cm = 10 mm

Thus, 5 mm = \frac{5}{10} cm = 0.5 cm

Therefore, 0.5 cm indicates 2.5 km

- Divide both by 0.5

Therefore, 1 cm represents 5 km

5 mm to 2.5 km corresponds to 1 cm to 5 km

E.

Since 1 mm signifies 500 m

- Convert mm to cm

Since 1 cm = 10 mm

So, 1 mm = \frac{1}{10} cm = 0.1 cm

- Convert m to km

Since 1 km = 1000 m

Thus, 1 m = \frac{1}{1000} = 0.001 km

Thus, 500 m = \frac{500}{1000} = 0.5 km

Thus, 0.1 cm indicates 0.5 km

- Multiply both by 10

Thus, 1 cm represents 5 km

1 mm to 500 m corresponds to 1 cm to 5 km

Learn more:

You can learn more about equivalent measurements in

3 0
3 days ago
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