find the length of the hypotenuse of a right triangle with legs of 12 ft and 16 ft. A 11 ft. B 18 ft. C 20 ft. D 28 ft

Mathematics

1 answer:

9966 [12]3 years ago

3 0

C 20ft, shoukd be the answer.

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∆ABC ~∆DEF, ∆ABC has a heights of 20 inches, and ∆DEF has a height of 24 inches. What is the ratio of the area of ∆ABC to the ar

OLEGan [10]

Answer:

The ratio of the area of ∆ABC to the area of ∆DEF is 25/36

Step-by-step explanation:

∆ABC ~∆DEF

∆ABC has a heights of 20 inches, and ∆DEF has a height of 24 inches

The ratio of the height (h1) of ∆ABC to the height (h2) of ∆DEF is:

h1 / h2= (20 inches) / (24 inches)

h1 / h2= 20 / 24

Simplifying the fraction dividing the numerator and the denominator by 4:

h1 / h2= (20/4) / (24/4)

h1 / h2= 5 / 6

The ratio of the area (A1) of ∆ABC to the area (A2) of ∆DEF is:

A1 / A2 = (h1 / h2)^2

A1 / A2 = (5 / 6)^2

A1 / A2 = 5^2 / 6^2

A1 / A2 = 25 / 36

5 0

3 years ago

Find the value of k for which the given function is a probability density function. f(x = kekx on [0, 2]

liraira [26]

Reading this as

$f_X(x)=\begin{cases}ke^{kx}&\text{for }x\in[0,2]\\0&\text{otherwise}\end{cases}$

For $f_X(x)$ to be a valid PDF, the integral over its support must equal 1:

$\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=\int_{x=0}^{x=2} ke^{kx}\,\mathrm dx=1$

Let $y=kx$ , so that $\dfrac{\mathrm dy}k=\mathrm dx$ and the integral becomes

$\displaystyle\int_{y=0}^{y=2k}ke^y\,\dfrac{\mathrm dy}k=\int_0^{2k}e^y\,\mathrm dy=e^y\bigg|_{y=0}^{y=2k}=1$
$e^{2k}-e^0=1$
$e^{2k}-1=1$
$e^{2k}=2$
$\ln e^{2k}=\ln2$
$2k=\ln 2$
$k=\dfrac{\ln 2}2=\ln\sqrt2$