Answer:
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Answer:
Step-by-step explanation:
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by
T(x) = 160-0.05x^2
a. [0, 10]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
The average temperature
= (160 + 155)/2 = 157.5
b. [10, 40]
For x = 10
T(10) = 160 - 0.05 × 10^2
T(10) = 160 - 5 = 155
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (80 + 155)/2 = 117.5
c. [0, 40]
For x = 0
T(0) = 160 - 0.05 × 0^2
T(0) = 160
For x = 40
T(10) = 160 - 0.05 × 40^2
T(10) = 160 - 80 = 80
The average temperature
= (160 + 80)/2 = 120
For rhombus A:
base = 7 in
area = 35 in²
Area of rhombus = b * h
⇒ 35 = 7 * h
⇒ h = 35/7
⇒ h = 5 in
Rhombus B:
height = 3*5 = 15 in
base = 7*3 = 21 in
Area = 15 * 21 = 315 in²
Area
of rhombus B is 9 times area of rhombus A. When the dimensions are
increased to 3 times the initial dimension, area became 9 times.
<h3>
Answer: $130</h3>
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Explanation:
I'm assuming the bank is using simple interest (instead of compound interest).
If so, then,
i = P*r*t
i = 8000*0.0325*0.5
i = 130
The amount of interest is $130
In the formula above, I used
- P = 8000 = amount deposited
- r = 0.0325 = annual interest rate in decimal form
- t = 6/12 = 0.5 years, which means you don't use t = 6. We don't use t = 6 because the interest rate is on an annual basis, and not a 6-month basis.
