Let AB extended intersect DC extended at point E
<span>We now have right triangle BEC with E = 90 degrees </span>
<span>For triangle BEC: </span>
<span>Exterior angle at E = 90 </span>
<span>Exterior angle at C = 148 (given) </span>
<span>Exterior angle of all polygons add up to 360 degrees </span>
<span>Exterior angle at B = 360−148−90 = 122 </span>
<span>So in quadrilateral ABCD </span>
<span>B = 122 </span>
<span>D = 360−44−148−122 = 46</span>
Answer:
→ The table is:
→ x → -1 → 0 → 1
→ y → -3 → 0 → 3
The graph of the line is figure d
Step-by-step explanation:
∵ y = 3x
∵ x = -1, 0, 1
→ Substitute the values of x in the equation to find the values of y
∴ y = 3(-1) = -3
∴ y = 3(0) = 0
∴ y = 3(1) = 3
→ The table is:
→ x → -1 → 0 → 1
→ y → -3 → 0 → 3
∵ The form of the linear equation is y = m x + b, where
∵ y = 3x
→ Compare the equation with the form
∴ m = 3
∴ b = 0
→ That means the slope is positive, then the direction of the line must
be from left tp right and passes through the origin
∴ The graph of the line is figure d
The top part equals 1 because anything to the power of zero equals one. then for the second one bring every variable with a negative exponent to the top and leave the 2 at the bottom
Answer:
Given
Step-by-step explanation:
Given that: △RST ~ △VWX, TU is the altitude of △RST, and XY is the altitude of △VWX.
Comparing △RST and △VWX;
TU ~ XY (given altitudes of the triangles)
<TUS = <XYW (all right angles are congruent)
<UTS ≅ <YXW (angle property of similar triangles)
Thus;
ΔTUS ≅ ΔXYW (congruent property of similar triangles)
<UTS + <TUS + < UST = <YXW + <XTW + <XWY = 
(sum of angles in a triangle)
Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX
So that:
= 
(corresponding side length proportion)
