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3 years ago

What is a differential equation

1 answer:

5 0

A differential equation is one that contains the derivative of a function. For example f(x) + 3 = 4f'(x) - 2. Usually you will be given one of the functions and initial conditions if you have to solve.

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The legs of the isosceles triangle each measure 14 inches. calculate the length of the hypotenuse

Evgen [1.6K]

Answer:

I'm guessing you are talking about an isosceles RIGHT triangle.

In such a triangle, the length of the hypotenuse equals the length of either leg times the square root of 2.

hypotenuse = 14 * square root of 2 =

14 * 1.4142135623731 = 19.79898987322330

or rounding that equals 19.80 inches

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Allometric relations often can be modeled by f(x)= x^b, where a and b are constants. One study showed that for a male fiddler

dolphi86 [110]

Answer:

a) $f(2)= 0.448 (2)^{1.21}= 1.036 grams$

b) $0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$

Step-by-step explanation:

For this case we know the following function:

$f(x) = 0.448 x^{1.21}$

The notation is: x is the weight of the crab in grams, and the output f(x) is the weight of the claws in grams.

Part a

For this case we just need to replace x = 2 gram in the function and we got:

$f(2)= 0.448 (2)^{1.21}= 1.036 grams$

Part b

For this case we know tha value for $f(x) =0.5$ and we want to find the value of x who satisfy this condition:

$0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$

8 0

3 years ago

Q: Marci wants to prove that the diagonals of

Trava [24]

Answer: Option B

She should use the fact that the

opposite sides of a parallelogram are

congruent and then use the

Pythagorean theorem.

We cannot use the diagonals of square property because this is a rectangle, opposite angles will also not work, and we cannot use the diagonal property because thats what we have to prove.

Must click thanks and mark brainliest

7 0

3 years ago

A cylinder has a height of 23 m and a volume of 18,488 m³. What is the radius of the cylinder? Round your answer to the nearest

diamong [38]

The radius to this cylinder is 16.

3 0

3 years ago

Read 2 more answers

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.8-6}{0.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5-6}{0.2}$

$Z = -5$

$Z = -5$ has a pvalue of 0.

1 - 0 = 1

5 0

3 years ago