we know that
The measure of the external angle is the semidifference of the arcs that it covers
so
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
Let
x------> minor arc MN
major arc MN=360-x
substitute in the formula above
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
75.86=(1/2)*(360-x-x)------> 151.72=(360-2x)------>360-151.72=2x
x=104.14°
therefore
the answer is
the measure of arc MN (minor arc) is 104.14°
Answer:
C. 3 in x 6 in
Step-by-step explanation:
Jammal cuts the block in a straight line parallel to one side... so the section revealed when he finishes his cut will be identical as the parallel side to which the cut is done.
We know the the left side of the prism on the image is 3 inches wide and 6 inches high... so that will also be the dimensions of the exposed cross section.
The answer is then 3 inches y 6 inches. The thickness of the block (5 inches) has no impact on the exposed area of the cross-section.
Answer:
-0.04
Step-by-step explanation:
4.23-5= -0.77
-0.77*0.04= -0.03
-0.77*0.09= -0.07
-0.07 + -0.03= -0.04
Step 1: Simplify both sides of the equation.
15.5=26−7p
15.5=−7p+26
Step 2: Flip the equation.
−7p+26=15.5
Step 3: Subtract 26 from both sides.
−7p+26=15.5
-26 -26
−7p=−10.5
Step 4: Divide both sides by -7.
-7p = -10.5
/-7 /-7
Answer:
P = $1.5 (Price for each game)
Hope this helps!
~LENA~
C)!!!! Probably too late? ;)
