Answer:
B
Step-by-step explanation:
Let point N has coordinates (0,0), then figure A has vertices at points (-3,0), (-7,0), (-7,4) and (-3,4). Figure B has vertices at points (3,1), (7,1), (7,-3) and (3,-3).
1. Translation 10 units to the right has the rule
(x,y)→(x+10,y).
Then vertices of the figure A have images:
- (-3,0)→(7,0);
- (-7,0)→(3,0);
- (-7,4)→(3,4);
- (-3,4)→(7,4).
2. Translation 3 units down has the rule
(x,y)→(x,y-3).
Then images are
- (7,0)→(7,-3);
- (3,0)→(3,-3);
- (3,4)→(3,1);
- (7,4)→(7,1).
As you can see these points are exactly the vertices of the figure B.
Answer:
x is 46.67
y is 28.4°
Step-by-step explanation:
To solve for x we use sine
sin ∅ = opposite / hypotenuse
AC is the opposite
x is the hypotenuse
so we have
sin 59° = 40 / x
x sin 59 = 40
Divide both sides by sin 59
x = 40/sin 59
x = 46.67
To solve for y we also use sine
sin y = 30/ 63
y = sin^-1(30/63)
y = 28.4°
Hope this helps you
The airplane has descended (25,000 - 19,000) = 6,000 feet
while flying (150 - 90) = 60 miles.
If the descent is modeled by a linear function, then the slope
of the function is
(-6000 ft) / (60 miles) = - 100 ft/mile .
Since it still has 19,000 ft left to descend, at the rate of 100 ft/mi,
it still needs to fly
(19,000 ft) / (100 ft/mile) = 190 miles
to reach the ground.
It's located 90 miles west of the runway now. So if it continues
on the same slope, it'll be 100 miles past the runway (east of it)
when it touches down.
I sure hope there's another airport there.