Answer:
82 years and 3 quarters
Step-by-step explanation:
A = P (1 + r / n)ⁿˣ
P = Principal amount
r = Annual interest rate
n = Number of compounds per year
x = time in years
A = Amount after time 'x'
10500 = 2100 (1 + 0.0195 / 4)⁴ˣ
Divide the whole equation by 2100
10500 / 2100 = ({2100 (1 + 0.004875)} / 2100 )⁴ˣ
5 = (1.004875)⁴ˣ
Taking Natural logarithm (㏑) on both sides
㏑ 5 = ㏑ (1.004875)⁴ˣ
㏑ 5 = 4x ㏑ (1.004875)
1.6094 = 4x (0.004863)
1.6094 = 0.01945x
x = 82.75
So, If compounded quarterly at an APR of 1.95% the amount deposited in savings account of $2100 will accumulate to $10500 in 82 years and 3 quarters.
Let the number = x
You have 15 times the number so that is 15x
9 less would be subtracting 9:
15x -9
Is 6 means the equation equals 6:
The equation is: 15x-9 = 6
Answer:
c
Step-by-step explanation:
Step 1
Solve 3x-y=-63x−y=−6 for yy.
y=6+3xy=6+3x
Step 2
Substitute 6+3x6+3x for yy in the second and third equation.
2\left(6+3x\right)-5z=-52(6+3x)−5z=−5 x-\left(6+3x\right)-2z=-4x−(6+3x)−2z=−4
Step 3
Solve these equations for xx and zz respectively.
x=-\frac{17}{6}+\frac{5}{6}zx=−
6
17
+
6
5
z z=-1-xz=−1−x
Step 4
Substitute -\frac{17}{6}+\frac{5}{6}z−
6
17
+
6
5
z for xx in the equation z=-1-xz=−1−x.
z=-1-\left(-\frac{17}{6}+\frac{5}{6}z\right)z=−1−(−
6
17
+
6
5
z)
Step 5
Solve z=-1-\left(-\frac{17}{6}+\frac{5}{6}z\right)z=−1−(−
6
17
+
6
5
z) for zz.
z=1z=1
Step 6
Substitute 11 for zz in the equation x=-\frac{17}{6}+\frac{5}{6}zx=−
6
17
+
6
5
z.
x=-\frac{17}{6}+\frac{5}{6}\times 1x=−
6
17
+
6
5
×1
Step 7
Calculate xx from x=-\frac{17}{6}+\frac{5}{6}\times 1x=−
6
17
+
6
5
×1.
x=-2x=−2
Step 8
Substitute -2−2 for xx in the equation y=6+3xy=6+3x.
y=6+3\left(-2\right)y=6+3(−2)
Step 9
Calculate yy from y=6+3\left(-2\right)y=6+3(−2).
y=0y=0
Step 10
The system is now solved.
x=-2x=−2 y=0y=0 z=1z=1
Solution
x=-2x=−2
y=0y=0
z=1z=1
Slope = (y2-y1)/(x2-x1)
Slope = (6 - 3)/(-4 - 2)
Slope = 3/-6
Slope = -1/2
Answer
<span>-1/2</span>
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