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3 years ago

Similar Triangles Question w/ Picture

Mathematics

1 answer:

melomori [17]3 years ago

7 0

Answer:

$EF=\dfrac{120}{7}\ ft$

Step-by-step explanation:

Triangles CEF and CAD are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so

$\dfrac{EF}{AD}=\dfrac{CF}{CD}\\ \\\dfrac{EF}{40}=\dfrac{CF}{CD}$

Triangles DEF and DBC are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so

$\dfrac{EF}{BC}=\dfrac{DF}{DC}\\ \\\dfrac{EF}{30}=\dfrac{CD-CF}{CD}\\ \\\dfrac{EF}{30}=1-\dfrac{CF}{CD}$

Substitute the fraction $\frac{CF}{CD}$ from the first equality into the second equality:

$\dfrac{EF}{30}=1-\dfrac{EF}{40}\ [\text{Muliply by 120}]\\ \\4EF=120-3EF\\ \\4EF+3EF=120\\ \\7EF=120\\ \\EF=\dfrac{120}{7}=17\dfrac{1}{7}\ ft$

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netineya [11]

Answer:

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Step-by-step explanation:

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A petrol kiosk p is 12 km due north of another petrol kiosk q. The bearing of a police station r from p is 135 degree and that f

tiny-mole [99]

Answer:

Distance between P and R is 40.15 km.

Step-by-step explanation:

From the picture attached,

Petrol kiosk P is 12 km due North of another petrol kiosk Q.

Bearing of a police station R is 135° from P and 120° from Q.

m∠QPR = 180° - 135° = 45°

m∠PQR = 120°

m∠PRQ = 180° - (m∠QPR +m∠PQR)

= 180° - (45° + 120°)

= 180° - 165°

= 15°

Now we apply sine rule in ΔPQR to measure the distance between P and R.

$\frac{\text{sin}(\angle QPR)}{\text{QR}}= \frac{\text{sin}(\angle PQR)}{\text{PR}}=\frac{\text{sin}\angle PRQ}{\text{PQ}}$

$\frac{\text{sin}(45)}{\text{QR}}= \frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}$

$\frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}$

PR = $\frac{12\text{sin}(120)}{\text{sin}(15)}$

PR = 40.15 km

Therefore, distance between P and R is 40.15 km.

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3 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

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Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

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A linear function has a y-intercept of -12 and a slope of 3/2 what’s is the equation of the lone

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Answer:

A.Y=3/2x-12

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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