The side lengths of triangle are 6 units, 8 units and 10 units.
<u>SOLUTION:
</u>
Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)
We know that, distance between two points 
is given by
Now,
9514 1404 393
Answer:
∠CAB = 28°
∠DAC = 64°
Step-by-step explanation:
What you do in each case is make use of the relationships you know about angles in a triangle and around parallel lines. You can also use the relationships you know about diagonals in a rectangle, and the triangles they create.
<u>Left</u>
Take advantage of the fact that ∆AEB is isosceles, so the angles at A and B in that triangle are the same. If we call that angle measure x, then we have the sum of angles in that triangle is ...
x + x + ∠AEB = 180°
2x = 180° -124° = 56°
x = 28°
The measure of angle CAB is 28°.
__
<u>Right</u>
Sides AD and BC are parallel, so diagonal AC can be considered a transversal. The two angles we're concerned with are alternate interior angles, so are congruent.
∠BCA = ∠DAC = 64°
The measure of angle DAC is 64°.
(Another way to look at this is that triangles BCE and DAE are congruent isosceles triangles, so corresponding angles are congruent.)
Answer:
V=4
/3 times pie(3.14) times r^3
Step-by-step explanation:
Answer:
it is 4,000
Step-by-step explanation:
since you are rounding it to the nearest thousand, you loo at the 4, and the 2, the 2 is blow five, so it doesn't change the thousands place, but everything else becomes a 0
Ok, let’s take it step by step.
In order to write an equation in slope intercept form, solve for y.
3x+4y=12
First, subtract 3x from both sides.
4y=-3x+12
Now, divide both sides by 4.
y=-3/4x+3.
So, 3x+4y=12 written in slope intercept form is y=-3/4x+3.
