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3 years ago

Consider the figure below. Find the values of J, W, and A

Mathematics

1 answer:

stira [4]3 years ago

6 0

Answer/Step-by-step explanation:

✔️Find W using the right triangle altitude theorem:

Thus,

W = √(15/2 × 10)

W = √75

W = 8.7 (nearest tenth)

✔️Find J using Pythagorean Theorem:

J² = W² + (15/2)²

J² = 8.7² + (15/2)²

J² = 131.94

J = √131.94

J = 11.5 (nearest tenth)

✔️Find A using Pythagorean Theorem:

A² = W² + (10)²

A² = 8.7² + (10)²

A² = 175.69

A = √175.69

A = 13.3 (nearest tenth)

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HURRY. Write a ratio for the following word phrase. Write the fraction in lowest terms.

mestny [16]

Answer:

84:2 42/1

Step-by-step explanation:

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3 years ago

angelo rode his bike around a bike trail that was 1/4 of a mile long.he rode his bike around thetrail 8 times.Angelo says he rod

forsale [732]

Hi there!

From this problem we can pull out some key information and remove any excess.

- The trail is 1/4 of a mile long.
- Angelo rode around the trail 8 times.
- Angelo claims he rode 8/4 miles.
- Teresa claims he rode 2 miles.

From that we know that if Angelo rode around the trail 8 times and the trail is 1/4 a mile, he in total rode 8/4 of a mile.

However, Teresa claims he rode 2 miles!

And the answer is... they're both correct!

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3 years ago

A culture is grown with a population of 45 million cells. The cell population is found to double every hour. What will the popul

Alexandra [31]

45 million * $2^{4}$ = 45million * 16 = 720million

Or you can use
$y=Ae^{kt}$
A is equal to 45 million
To find the rate, k, we can plug in:
$90million=45million*e^{k*1}
$
$2=e^{k}$
$k=ln(2)$
$y=4.5*10^{7}*e^{t *ln2}$
Knowing that t = 4
$y = 4.5*10^{7}*e^{ln(2)*4}
$
$y = 720 million$

6 0

3 years ago

Read 2 more answers

Find the area of the circle with the given radius or diameter. Use = 3.14. d=13

leva [86]

Area of a triangle = $\pi r^{2}$
r = diameter/2
so area = pi * (13/2)^2
= 3.14*6.5*6.5 = 132.665

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3 years ago

Calculate the average travel time for each distance, and then use the results to calculate. A 6-column table with 3 rows in the

BigorU [14]

The average travel time is simply the mean travel time between the given points

The average time that it takes for the car to travel the first 0.25m is 2.23s.
The average time to travel just between 0.25m and 0.50m is 0.9s.
Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.

(a) The average time to travel the first 0.25m

The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.

So, the average time to travel is:

$Time = \frac{2.24s + 2.21s + 2.23s}{3}$