3 + 0.15x = 0.25x
3 = 0.10x
30 = 1x
The given equation is
By solving the given equation, we get
While solving the equation,we get 0=0.
Therefore, the given equation is true for all the values of y.
Hence, we can say that the given equation has infinite number of solutions.
Answer:
15 pieces of string all 2 ft long
Step-by-step explanation:
12 + 18 ft = 30 ft
30 / 2 = 15
15 pieces of string
2 ft
Two ways to answer:
A. Using calculus:
H(t)=-16t^2+65t+3
To find maximum of H(t), equate H'(t)=0, or
H'(t)=-32t+65=0 => t=65/32=2.03125, or
t=2.0 s. to the nearest 10
B, Without using calculus, i.e. by completing the square
H(t)
=-16t^2+65t+3
=-16(t^2-65/16t)+3
=-16(t^2-2*(65/32)t+(65/32)^2)+3+(65/32)^2
=-16(t-65/32)^2+3+(65/32)^2
meaning the maximum occurs at (t-65/32)=0, or t=65/32=2.03125
Answer: t=2.0 seconds (to the nearest tenth)
Answer:
The measure of ∠GKH is 27°
Step-by-step explanation:
- In the isosceles triangle, the base angles are equal in measures
- The measure of an exterior angle at a vertex of a triangle equals the sum of the measures of two opposite interior angles
In Δ HJK
∵ HJ = JK
→ That means the triangle is isosceles
∴ Δ HJK is an isosceles triangle
∵ ∠JHK and ∠JKH are base angles
→ By using the first rule above
∴ m∠JHK = m∠JKH
∵ m∠HJK = 70°
∵ m∠JHK + m∠JKH + m∠HJK = 180° ⇒ interior angles of a triangle
∴ m∠JHK + m∠JKH + 70 =180
→ Subtract 70 from both sides
∴ m∠JHK + m∠JKH = 110
→ Divide their sum by 2 to find the measure of each one
∴ m∠JHK = m∠JKH = 110 ÷ 2 = 55°
∵ ∠JHK is an exterior angle of ΔGHK at vertex H
∵ ∠HGK and ∠GKH are the opposite interior angles to ∠JHK
→ By using the 2nd rule above
∴ m∠JHK = m∠HGK + m∠GKH
∵ m∠JHK = 55°
∵ m∠HGK = 28°
∴ 55 = 28 + m∠GKH
→ Subtract 28 from both sides
∴ 27° = m∠GKH
∴ The measure of ∠GKH is 27°