A. 305
b. 900
c. 46
d. 157,019
<u>Answer</u><u> </u><u>:</u><u>-</u>
9(3+√3) feet
<u>Step </u><u>by</u><u> step</u><u> explanation</u><u> </u><u>:</u><u>-</u>
A triangle is given to us. In which one angle is 30° and length of one side is 18ft ( hypontenuse) .So here we can use trignometric Ratios to find values of rest sides. Let's lable the figure as ∆ABC .
Now here the other angle will be = (90°-30°)=60° .
<u>In ∆ABC , </u>
=> sin 30 ° = AB / AC
=> 1/2 = AB / 18ft
=> AB = 18ft/2
=> AB = 9ft .
<u>Again</u><u> </u><u>In</u><u> </u><u>∆</u><u> </u><u>ABC</u><u> </u><u>,</u><u> </u>
=> cos 30° = BC / AC
=> √3/2 = BC / 18ft
=> BC = 18 * √3/2 ft
=> BC = 9√3 ft .
Hence the perimeter will be equal to the sum of all sides = ( 18 + 9 + 9√3 ) ft = 27 + 9√3 ft = 9(3+√3) ft .
<h3>
<u>Hence </u><u>the</u><u> </u><u>perim</u><u>eter</u><u> of</u><u> the</u><u> </u><u>triangular</u><u> </u><u>pathway</u><u> </u><u>shown</u><u> </u><u>is</u><u> </u><u>9</u><u> </u><u>(</u><u> </u><u>3</u><u> </u><u>+</u><u> </u><u>√</u><u>3</u><u> </u><u>)</u><u> </u><u>ft</u><u> </u><u>.</u></h3>
Answer:
11- All 0's are on the x and y planes
12- All positive points/numbers are where the x and y axis are both positive (Quadrant 1, 2, and 4)
13- All negative points/numbers would be wherever the x and y axis are negative (Quadrants 2, 3, and 4)
Answer:
Step-by-step explanation:
The line has a slope of -1/3. For lines to be perpendicular there product must equal negative 1
m1(m2)=-1
-m/3=-1
m=3
So we know our line has a slope of three so
y=3x+b, we can solve for the y intercept, b, using point (-2,8)
8=3(-2)+b
8=-6+b
b=14, so our line is
y=3x+14
How about this (see attached image):
Use the four cuts as shown in the image (red lines).
Then assemble 5 equal squares by the numbers: 1 center square and the rest are pieced together using two pieces as shown. All five together add up to the same area as original square because we use all pieces.
The way one gets a hint toward a solution is to see how an area of a square of length 1 can be split into 5 equal square areas:

which indicates we need to find a a triangle with sides 2 and 1 to get the hypotenuse of the right length. That gave rise to the cut pattern (if you look carefully, there are triangles with those side lengths).