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Troyanec [42]
3 years ago
7

There is a triangular parking lot at the local mall. The second angle of the triangular parking lot is six more than twice as la

rge as the first angle. The third angle is equal to the sum of the other two angles. What are the measures of the three angles?
Mathematics
1 answer:
Travka [436]3 years ago
8 0

Answer:

The measure of the three angles are;

The first angle is 62°

The second angle is 28°

The third angle is 90°

Step-by-step explanation:

The given information are;

The shape of the parking lot = Triangular

Let the angles of the triangle be given as follows

First angle = A

Second angle = B

Third angle = C

The given triangle interior angle dimensions are;

A = 6 + 2× B

C = A + B

However, we have;

A + B + C = 180° (Angle sum property for a triangle)

Therefore;

A + B + C = 180° gives;

C + C = 180° (Transitive property)

2·C = 180°

C = 180°/2 = 90°

C = 90°

However, C = A + B  therefore;

90° = A + B and, A = 6 + 2 × B, we get;

A + B = 90° (Symmetric property)

6 + 2× B + B = 90° (Substitution property)

6 + 3·B = 90°

3·B = 90° - 6° = 84°

B = 84°/3 = 28°

B = 28°

From, A = 6 + 2 × B, we have;

A = 6 + 2 × 28° = 62°

A = 62°

First angle = A = 62°

Second angle = B = 28°

Third angle = C = 90°

The measure of the three angles are;

First angle is 62°

Second angle is 28°

Third angle is 90°.

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11th grade geometry:
Aliun [14]

Answer:

<em>The perimeter is 72 units and the area is 149 square units.</em>

Step-by-step explanation:

\triangle SBA has coordinates S(15,-8),B(-2,21) and A(0,0)

Using the distance formula.........

Length of side SB = \sqrt{(15+2)^2+(-8-21)^2}= \sqrt{17^2+(-29)^2}= \sqrt{1130}

Length of side BA= \sqrt{(-2)^2+(21)^2}= \sqrt{445}

Length of side AS =\sqrt{(15)^2+(-8)^2}=\sqrt{289}=17

So, the perimeter of the triangle will be:  (SB+BA+AS)= \sqrt{1130}+ \sqrt{445}+17 =71.71... \approx 72 units.   <em>(Rounded to the nearest unit)</em>

The height of the triangle for the corresponding base SB is 8.89 units.

<u>Formula for the Area of triangle</u>,  A= \frac{1}{2}(base\times height)

So, the area of the \triangle SBA will be:  \frac{1}{2}(\sqrt{1130}\times 8.89)= 149.42... \approx 149 square units.   <em>(Rounded to the nearest unit)</em>

3 0
3 years ago
This spinner is spun 36 times. The spinner landed on A 6 times, on B 21 times, and on C 9 times. Compute the empirical probabili
Kruka [31]

Answer: 7/12

Step-by-step explanation:

number of times it landed on A = 6

number of times it landed on B = 21

number of times it landed on C = 9

Total number = 36

The empirical probability that the spinner will land on B is given by

P(B) = number of times it landed on B / Total number , that is

p(B) = 21/36

P(B) =7/12

Note: Empirical means verifiable by observation or experience rather than theory and it was verified that it landed on B 21 times.

6 0
3 years ago
I need help please.
Sidana [21]

Answer:

-1/63y^11

Step-by-step explanation:

because we multiply it

7 0
3 years ago
Find the common ratio and the explicit formula <br><br>-3, 9, -27, 81,​
Fofino [41]

Answer:

see explanation

Step-by-step explanation:

These are the terms of a geometric sequence with n th term

a_{n}\\ = a(r)^{n-1}

where re a is the first term and r the common ratio

r = 9 ÷ - 3 = - 27 ÷ 9 = 81 ÷ - 27 = - 3 ← common ratio

and a = - 3, thus

[te x]a_{n}[/tex] = - 3(-3)^{n-1} ← explicit formula

6 0
3 years ago
Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height ov
Semmy [17]

Answer: p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

3 0
3 years ago
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