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3 years ago

Solve the following equation.

Mathematics

1 answer:

vitfil [10]3 years ago

4 0

Answer:

x=10/3=3.333 or = - 10/3=-3.333

Step-by-step explanation:

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A surveyor is measuring the distance across a canyon. He sets up two similar right triangles and finds that the length of CD = 1

Citrus2011 [14]

The length side AB is 64 ft , Option D is the correct answer.

<h3>What is a Right triangle ?</h3>

A triangle that has one angle 90 degree is called Right angled Triangle.

From the figure it can be seen that

The triangle CDE is a right Triangle

Applying Pythagoras theorem to find the length of CE

CE² = 12² + 16²

CE = 20 ft

As both the triangles are similar therefore the ratio of the sides will be equal

BC : CD = AC : CE

48 / 12 = AC / 20

AC = 80ft

As Triangle ABC is also a right angled triangle

Applying Pythagoras theorem to find the length of AB

AC² = BC² +AB²

6400 = 48² +AB²

AB² = 4096

AB = 64 ft

Therefore Option D is the correct answer.

To know more about Right Triangle

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3 0

1 year ago

Plz help I have no idea how to answer it

saveliy_v [14]

5. Answer: see explanation

<u>Step-by-step explanation:</u>

If the roots are m and 3m, then x = m and x = 3m

⇒ x - m = 0 and x - 3m = 0

⇒ (x - m)(x - 3m) = 0

⇒ x² - 4mx + 3m² = 0

Since x² + px + q = 0 then p = -4m and q = 3m²

3p² = 3(-4m)² 16q = 16(3m²)

= 3(16m²) = 48m²

= 48m²

3p² = 48m² = 16q ⇒ 3p² = 16q

**********************************************************************************************

6. Answer: 8 or 18

<u>Step-by-step explanation:</u>

The Area of the entire rectangle (A = L × w) is 12 × 10 = 120

The Area of the shaded region is 72, so the Area of the non-shaded region is 120 - 72 = 48.

There are two non-shaded triangles.

Bigger non-shaded Δ: L = 12-x, w = 10 ⇒ $A = \dfrac{10(12-x)}{2}$
Smaller non-shaded Δ: L = x, w = x ⇒ $A=\dfrac{x(x)}{2}$

Combine the Areas of both triangles and set it equal to the Area of the non-shaded region:

$\dfrac{x(x)}{2}+\dfrac{10(12-x)}{2}=48\\\\\\\dfrac{x^2}{2}+\dfrac{120-10x}{2}=48\\\\\\\dfrac{x^2-10x+120}{2}=48\\\\\\x^2-10x+120=96\\\\x^2-10x+24=0\\\\(x-4)(x-6)=0\\\\x - 4 = 0\quad or\quad x-6=0\\x=4\qquad or\qquad x=6\\\\\text{Based on the image provided, it appears that the x-value will be the}\\\text{smallest of the two possible solutions (4), but both solutions are valid.}$