We develop two equations based on the given diagram. We let x be the length from C to the point meeting the endpoint of 5. We let y be the length of CD.
First, we determine the length from D to the endpoint of line measuring 9 cm. We use the Pythagorean theorem.
l = sqrt ((3)² + 5²)
l = 5.83
Then, the equations that can be developed or established are:
x² + 6² = y²
9² + (5 + x)² = (y + 5.83)²
In solving for x and y, we use substitution.
From the first equation,
y = sqrt (x² + 36)
We substitute this to the second equation y.
The value of x is 9.51.
y = sqrt (9.51² + 36) = 11.25.
Calculate for the area of the bigger triangle.
A = 0.5(9 cm)(5 + 9.51 cm) = 65.295 cm²
Also calculate for the area of the smaller triangle,
A = (0.5)(6 cm)(9.51 cm) = 28.53 cm²
The difference between the areas is 36.765 cm².
<em>ANSWER: 36.77 cm²</em>
Answer:
(20, -2)
Step-by-step explanation:
Given the coordinate vectors p=(4,3) and q=(-2,7),
4p - 2q = 4(4, 3)- 2(-2, 7)
4p - 2q = (16, 12) - (-4, 14)
4p - 2q = (16+4, 12-14)
4p - 2q =(20, -2)
hence the required vector is (20, -2)
6,321,457....rounded to the nearest hundred thousand = 6,300,000
Answer:
6 liters of lemon-lime soda
3 pints of sherbet
9 cups of ice
Step-by-step explanation:
divide the original by 2 the add that to the original
Answer:
Step-by-step explanation:
<u>Solving</u>
- 3tanx = -4
- tanx = -4/3
- tan values are negative in the 2nd and 4th quadrant
- The base value is x = 53°
In <u>Quadrant II</u>,
- 90° + 53°
- ⇒ <u>x = 143°</u>
In <u>Quadrant IV</u>,
- 270° + 53°
- ⇒ <u>x = 323°</u>