Answer:
5 km
Step-by-step explanation:
From. The question, the triangle formed us a right angled triangle,
h km directly from starting point is the hypotenus, h :
Hypotenus = sqrt(opposite² + Adjacent²)
h = sqrt(4² + 3²)
h = sqrt(16 + 9)
h = sqrt(25)
h = 5 km
It's kinda like using variables, but shapes in their place. Let's set the circle to x, the triangle to y, and the square to z. Now the equations say x+(y+z)=13, (y+z)=8, and x+y=7. First thing we can do it plug 8 into the first equation for y+z, getting x+8=13. Subtract 8 to both sides and x (or circle) equals 5. Plug 5 into the last equation, x+y=7 to solve for y, getting 5+y=7. Subtract 5 to both sides and y (or triangle) equals 2. Finally, plug two in for y in y+z=8, getting 2+z=8. Subtract 2 from both sides and z (or square) equals 6. Hope I've explained it clearly!!
Answer:
Attached diagram A'B'C'D'
Step-by-step explanation:
Given is a quadrilateral ABCD. It says to draw a dilated version with a scale factor 2/3.
We see that scale factor is less than 1 which means it shrinks the image to a smaller one.
To draw a scaled copy, we need to find the lengths of its sides.
To do so, we can draw the diagonals AC & BD, and they intersect at origin O(0,0) such that OA= -2, OB= 2, OC= 4, OD= -4.
Applying a scale factor of 2/3, we get OA' = -4/3, OB' = 4/3, OC' = 8/3, OD' = -8/3.
So we have attached a scaled copy A'B'C'D' of quadrilateral ABCD with a scale factor 2/3.
Answer:
104°
Step-by-step explanation:
If segments NO and NM are congruent, then angles NMO and NOM are congruent. So, their supplements, angles NML and NOP are congruent. That is ...
∠NML ≅ ∠NOP = 104°
∠NML = 104°
