The answer would be 100/3
Answer:
Amount pay after one year for compounded quarterly = Rs 5627.54
Step-by-step explanation:
Given as,
Manu took loan of Rs 5000 , So, Principal = Rs 5000
The rate of interest applied = 12% per annum compounded quarterly
The loan took for period of year = one
Now from the compounded method :
For compounded quarterly
Amount = principal 
Or, Amount = Rs 5000 
Or, Amount = 5000 
Or, Amount = 5000 × 1.1255
∴ Amount = Rs 5627.54
Hence , The amount which Manu pay after one year at 12% per annum compounded quarterly is Rs 5627.54 Answer
The transformed function is G(x) = -4x² after applying the transformation stretched vertically and flipped over the x-axis option (C) G(x) = -4x² is correct.
<h3>What is a function?</h3>
It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
The options are missing.
The options are:
A. G(x) = 4x²
B. G(x) = -(1/4)x²
C. G(x) = -4x²
D. G(x) = (1/4)x²
We have an equation of a function F(x)
F(x) = x²
The transformation F(x) can be stretched vertically and flipped over the x-axis to produce the graph of G(x)
To stretch vertically if the function is multiplied by a constant value
f(x) = ax²
To flip over the x-axis if multiply by negative value.
g(x) = -ax²
From the options
G(x) = -4x²
Thus, the transformed function is G(x) = -4x² after applying the transformation stretched vertically and flipped over the x-axis option (C) G(x) = -4x² is correct.
Learn more about the function here:
brainly.com/question/5245372
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Answer:
am i suppoesd to do number 7 as well?
Step-by-step explanation:
Two wires
tether a balloon to the ground, as shown. How
high is the balloon above the ground? (Must Use Right
Triangle Trigonometry)
look at picture
Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8