24 / (3/8) =
24 * 8/3 =
192/3 =
64......there are 64 three-eights quart jars
Step 1: Find the total area. Since it is a square all the sides are 12 ft so multiply 12 by 12
12 x 12 = 144
^^^This is the total area including the area around the pool. We just need the are of the pool
Step 2: Outside of the pool there are four little 2 ft by 2 ft squares that aren't part of pool, and therefore won't need to be covered. The area of these 2 ft by 2 ft square is 4ft (2 times 2 is 4). Since there are 4 of these squares we multiply 4 by 4 to get 16. This means that 16 ft of the 144 ft we found previously are not part of the pool and we must subtract 16 from 144 to get the total pool area
144 - 16 = 128
This means that the area of the pool is 128 ft
Hope this helped and made sense! Let me know!
Answer:
y= 14
Step-by-step explanation:
Given that AC= BC, △ABC is an isosceles triangle.
∠CBA= ∠CAB (base ∠s of isos. △)
∠CBA= 50°
∠ACB +50° +50°= 180° (∠ sum of △ABC)
∠ACB +100°= 180°
∠ACB= 180° -100°
∠ACB= 80°
Given that ∠ACB= (5y +10)°,
(5y +10)°= 80°
5y +10= 80
5y= 80 -10 <em>(</em><em>-10</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
5y= 70 <em>(</em><em>simplify</em><em>)</em>
y= 70 ÷5 <em>(</em><em>÷</em><em>5</em><em> </em><em>throughout</em><em>)</em>
y= 14
<u>Answer:
</u>
x = 5 and y = 0 is correct solution of 2x + y -10 = 0 and x – y – 5 =0
<u>Solution:
</u>
Two given equations which needs to be solve are
2x + y – 10 = 0 ------ (1)
x– y – 5 = 0 ------ (2)
Let’s modify equation (1)
2x + y – 10 = 0
y =10 - 2x ------ (3)
On substituting value of y from equation (3) in equation (2) we get
x – (10 – 2x) -5 = 0
x – 10 + 2x – 5 = 0
3x -15 = 0
x = 5
Substituting x = 5 in equation (3) to get value of y.
y = 10 – 2
5 = 10 – 10 = 0
So on solving given equation we get x = 5 and y = 0.
Lets substitute value of x = 5 and y = 0 in equation (1) and equation (2) to check whether these calculated values satisfies given equations or not.
For equation (1), 2
5 + 0 – 10 = 10 – 10 = 0
For equation (2), 5 – 0 – 5 = 0
On solving, in both cases LHS = RHS for calculated values of x = 5 and y = 0.
Hence x = 5 and y = 0 is correct solution of two given equation.