Answer:D 237x/421x-515
Step-by-step explanation:
These are vertical angles so they are equal so
6a + 11 = 2a + 83
now solve for a
Answer: 1
Step-by-step explanation:
From the given picture, it can be seen that there is a plane H on which two pints J and K are located.
One of the Axiom in Euclid's geometry says that <em>"Through any given two points X and Y, only one and only one line can be drawn "</em>
Therefore by Axiom in Euclid's geometry , for the given points J and K in plane H , only one line can be drawn through points J and K.
<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:
600 m³
Step-by-step explanation:
Given:
- length of swimming pool = 30 m
- width of swimming pool = 10 m
- depth at deep end = 3 m
- depth at shallow end = 1 m
To find the volume of the water in the pool, find the area of a cross section (see attached image) and multiply it by the width of the pool.
<u>Area of cross section</u>
= area of rectangle - area of triangle
= (3 × 30) - [1/2 × 30 × (3 - 1)]
= 90 - 30
= 60 m²
<u>Volume of pool</u>
= area of cross section × width
= 60 × 10
= 600 m³
