pretty sure it would be 0.075
Answer:
one thousand, seven hundred twenty-eight
Answer:
-2x
Step-by-step explanation:
Graph
y > −2x + 3
Use the slope-intercept form to find the slope and y-intercept.
The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.
y = mx + b
Find the values of m and b using the form y = mx + b. m = −2
b = 3
The slope of the line is the value of m, and the y-intercept is the value of b.
Slope: −2
intercept: (0, 3)
Graph a dashed line, then shade the area above the boundary line since y is greater than
−2x + 3.
y > −2x + 3
S = πr(r + √(h² + r²))
400.2 = 3.14(6)(6 + √(h² + 6²))
400.2 = 18.84(6 + √(h² + 36))
18.84 18.84
21¹⁰⁹/₄₇₁ = 6 + √(h² + 36))
- 6 - 6
15¹⁰⁹/₄₇₁ = √(h² + 36)
231²²¹⁰⁰⁵/₂₂₁₈₄₁ = h² + 36
- 36 - 36
195²²¹⁰⁰⁵/₂₂₁₈₄₁ = h²
14 ≈ h
Answer:

Step-by-step explanation:
Given,
Principal ( P ) = $ 6000
Amount ( A ) = $ 14550
Time ( T ) = 10 years
Rate ( R ) = ?
<u>Finding </u><u>the </u><u>Interest</u>
The sum of principal and interest is called an amount.
From the definition,

plug the values
⇒
Swap the sides of the equation
⇒
Move 6000 to right hand side and change its sign
⇒
Subtract 6000 from 14550
⇒
Interest = $ 8550
<u>Finding </u><u>the </u><u>rate </u>

plug the values
⇒
Calculate
⇒
⇒
Hope I helped!
Best regards!!