(Please Help Asap) Evaluate 1/6 + 2/3 A. 1/5 B. 3/18 C. 5/6 D. 1 1/6

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

Travka [436]

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°.

8 0

2 years ago

Write the equation of the line that matches the scenario in slope intercept form.

viva [34]

Answer:

it took 60 minutes to fill up 800 gallons of water???

3 0

3 years ago

Need help don’t understand

adoni [48]

there is an equation for straight lines which goes

y-y,=m(x-x,)

where( x, , y, ) is any coordinate on the line and m is the gradient

so your answer is

y-7= -2(x-4)

does this make sense?

7 0

2 years ago

Read 2 more answers

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: