Answer:
When point A with coordinates (0, -1) is reflected across the x-axis and mapped onto point A', the coordinates of A' will be (0, 1).
i.e A'(0, 1) is the image of point A after a reflection.
Hence, point A is reflected across the x-axis.
Step-by-step explanation:
When we reflect a point A across the x-axis, the value of 'y' gets negated, but the value of 'x' remains unchanged.
In other words, when point P with coordinates (x, y) is reflected across the x-axis and mapped onto point P', the coordinates of P' will be (x, -y).
Thus, the rule is:
P(x, y) → P'(x, -y)
Thus, when point A with coordinates (0, -1) is reflected across the x-axis and mapped onto point A', the coordinates of A' will be (0, 1).
i.e A'(0, 1) is the image of point A after a reflection.
Hence, point A is reflected across the x-axis.
In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.
After k payments, the amount A still owed is
<span>A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i,
= (P-Mq/i)(1+[i/q])k + Mq/i.
</span>The amount of the fixed payment is determined by<span>M = Pi/[q(1-[1+(i/q)]-nq)].
</span>The amount of principal that can be paid off in n years is<span>P = M(1-[1+(i/q)]-nq)q/i.
</span>The number of years needed to pay off the loan isn = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
The total amount paid by the borrower is Mnq, and the total amount of interest paid is<span>I = Mnq - P.</span>
Answer:
Step-by-step explanation:
1,8
5,7
2,3
Answer:
For the first one use the formula
y=x*2
For the second use
y=x*4
The third is
y=x*15
The fourth is
y=x*16
Step-by-step explanation:
To find the value of the y units, u have to multiply any x unit with the first y unit given but not 0
idk if it makes sense but i tried
good luck!!
A)4^(n+3)=8^14
2^(2×(n+3))=2^(3×14)
2^(2n+6)=2^42
2^2n=2^36
n=18
b) (assuming a : is divide)
3^(2n+1)=9^17/3^3
3^(2n+1)=3^(2×16)/3^3
3^(2n+1)=3^29
3^2n=3^28
2n=28
n=14
d) (6^n)^4×36=216^10
6^4n×6^2=6^(3×10)
6^(4n+2)=6^30
6^4n=6^28
4n=28
n=7
e)7^(n^2)÷7=49^24
7^(n^2-1)=7^(2×24)
7^(n^2)=7^49
n^2=49
n=7
g)15^(n+4)÷5^(n+4)=81^6
3^(n+4)×5^(n+4)÷5^(n+4)=3^(4×6)
3^(n+4)=3^24
n=20
h)81^n÷9^n+9^(n+2)÷9=90÷9^6
9^2n÷9^n+9^(n+2)÷9=9*10/9^6
9^n+9^(n+1)=10/9^5
I don't know where to go from here
I)what?
