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Yakvenalex [24]
3 years ago
10

Look at the triangle on the top of the White House in the photo. Describe the size and angles of the triangle. Please Help me th

is is due tomorrow

Mathematics
1 answer:
vekshin13 years ago
5 0
You can use the size and angles to classify the triangle that is represented.

The peak is an obtuse angle, and the two vertices on the side are acute angles. This makes it an obtuse triangle. If there is one obtuse angle seen, it is classified as an obtuse triangle.

Two of the sides are the same length, and this makes it an isosceles triangle.
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( HELP? Will Mark :) )
Leya [2.2K]

Answer:

4

Step-by-step explanation:


3 0
3 years ago
Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and
Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

Then

y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0
3 years ago
Help plzzzzzzzzzzzzzzzz
Alex

Answer:

Last option.

Step-by-step explanation:

The consecutive exterior angle theorem states that the exterior angles are supplementary if two lines are cut by a transversal.

m∠1 + m∠7 = 180

m∠2 + m∠8 = 180

3 0
3 years ago
Trigonometry
Alchen [17]

The area of the triangle PQR is 17.6 square units.

Explanation:

Given that the sides of the triangle are PQ = 12 and PR = 3 and m\angle P = 78^{\circ}

We need to determine the area of the triangle PQR

<u>Area of the triangle:</u>

The area of the triangle can be determined using the formula,

\text {Area}=\frac{1}{2} qr \sin P

Substituting the values, we get,

\text {Area}=\frac{1}{2}(12)(3) \sin 78

Simplifying, we have,

\text {Area}=\frac{1}{2}(36)(0.98)

Multiplying the terms, we have,

\text {Area}=\frac{35.28}{2}

Dividing, we get,

\text {Area}=17.64

Rounding off to the nearest tenth, we have,

Area=17.6

Thus, the area of the triangle PQR is 17.6 square units.

3 0
3 years ago
Read 2 more answers
Please help me, I CANT fail this.
Annette [7]

Answer:

B

Step-by-step explanation:

4 0
3 years ago
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