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3 years ago

What is the inradius of a triangle with side lengths 9, 40, and 41?

Mathematics

1 answer:

Mars2501 [29]3 years ago

6 0

Answer: 20.5

Step-by-step explanation:

From the measure of the length of the sides of the triangle, 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9-40-41 is a Pythagorean triplet.

In a right angled triangle, the circumradius is half the hypotenuse.

In the given triangle, the hypotenuse = 41.

Therefore, the circumradius is 41÷2= 20.5 units.

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The price of a TV-set is $480. If it goes on sale at 18% off what will be the new price?

Alona [7]

Answer:

$393.60

Step-by-step explanation:

7 0

3 years ago

0.3f +1 = 0.2f - 4.5<br> What’s the answer step by step

jok3333 [9.3K]

Step-by-step explanation:

.3f+1 = .2f-4.5

1. Subtract .2f both sides

.1f+1 = -4.5

2. Subtract 1 from both sides

.1f = -5.5

3. Divide .1 both sides

f = -55

To check plug in -55 to all the f’s and both sides should be equal.

Hope this helps!

7 0

3 years ago

Read 2 more answers

Can anyone help me with this please!!

Levart [38]

Answer:

22.88 cm^3

Step-by-step explanation:

The volume of a sphere formula is (4/3)*pi * (radius)^3

Since the diameter is 5.08cm, we divide by 2 to find the radius which is 2.54 cm.

Also because of the fact that 2/3 is showing outside the cone, that means 1/3 is inside the cone. Therefore, we have to multiply the volume by 1/3.

We have:

Volume = (4/3) * pi * (2.54)^3 * (1/3)

= 22.88 cm^3

4 0

3 years ago

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

hoa [83]

Answer:

Obese people

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

$X \sim N(373,67)$

Where $\mu=373$ and $\sigma=67$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 373- 2(67) = 239$

$Upper = \mu + 2\sigma = 373+ 2(67) = 507$

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

$X \sim N(526,107)$

Where $\mu=526$ and $\sigma=107$

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

$Lower = \mu - 2\sigma = 526- 2(107) = 312$

$Upper = \mu + 2\sigma = 526+ 2(107) = 740$

The interval for the lean people is significantly higher than the interval for the obese people.

5 0

3 years ago