All categories

3 years ago

What does AY represent in this slope triangle? Ay Δx O Horizontal change O Slope angle T62

Mathematics

1 answer:

erastova [34]3 years ago

4 0

Answer:

Slope Angle. is the correct answer.

You might be interested in

I marked all 3 parts. I need to know part C (the shaded part). anybody mind helping?

Mnenie [13.5K]

Answer:

Area of shaded part ABCEF = 66 sq.cm

Step-by-step explanation:

AB = 8cm

CD = 8cm

Let DE = x cm

CE = 3x cm

CD = CE + DE = 8cm

x + 3x = 8

4x = 8

x = 8/4 = 2 cm

DE = 2cm

CE = 3 * 2 = 6 cm

Area of triangle ADE = 1/2 * base * height

= 1/2 * DE * AD

= 1/2 * 2 * 11 = 11 sq. cm

Area of triangle AEF = Area of triangle ADE = 11 sq. cm

Area of Rectangle ABCD = l * b = 8 * 11 = 88 sq.cm

Area of shaded part ABCEF = Area of Rectangle ABCD - (Area of triangle AEF + Area of triangle ADE)

= 88 - ( 11 + 11 ) = 88 -22 = 66 sq.cm

6 0

3 years ago

I'd really appreciate it if anyone helped! :) Giving Brainliest!

Anna11 [10]

the first three are irrational

5 0

3 years ago

3. Two cars are parked, 150m apart, and on opposite sides of a building. A camera on the top of the building rotates and can vie

allsm [11]

Answer:

<em>The building is 61.5 m tall</em>

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

h = height of the building

a, b = internal angles of each triangle

x = base of each triangle

The angles a and b can be easily found by subtracting the given angles from 90° since they are complementary angles, thus:

a = 90° - 37° = 53°

b = 90° - 42° = 48°

Now we apply the tangent ratio on both triangles separately:

$\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 53^\circ=\frac{150-x}{h}$

$\displaystyle \tan 48^\circ=\frac{x}{h}$

From the last equation:

$x=h.\tan 48^\circ$

Substituting into the first equation:

$\displaystyle \tan 53^\circ=\frac{150-h.\tan 48^\circ}{h}$

Operating on the right side:

$\displaystyle \tan 53^\circ=\frac{150}{h}-\tan 48^\circ$

Rearranging:

$\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}$

Solving for h:

$\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}$

Calculating:

h = 61.5 m

The building is 61.5 m tall

5 0

3 years ago

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

satela [25.4K]

Step-by-step explanation:

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - 5 + - \sqrt{ {5}^{2} - 4(5) } }{2} \\ x = \frac{ - 5 + - \sqrt{25 - 20} }{2} \\ x = \frac{ - 5 + - \sqrt{5} }{2}$

Therefore, the zeroes would be:

$x1 = \frac{ - 5 + \sqrt{5} }{2} \\ x2 = \frac{ - 5 - \sqrt{5} }{2}$

8 0

3 years ago

Reflect on the concept of function. What concepts (only the names) did you need to accommodate the concept of function in your m

disa [49]

The answer is idk cause idk

4 0

3 years ago