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jok3333 [9.3K]
3 years ago
14

Please help due IN MY CLASS TODAY

Mathematics
2 answers:
Lynna [10]3 years ago
6 0

Answer:

A. 3,4. E. Origin or 0,0 D. 6,-3

C. -2,-4 B. -3, 2

Step-by-step explanation:

Just go from x to y and I did this last year :p

valentinak56 [21]3 years ago
3 0
A(3,4) B(-3,2) C(-2,-4) D(6,-3)
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(-4,-6) and (3,-7) find the distance between each pair of points
Vlad [161]

Answer:

5\sqrt{2}

Step-by-step explanation:

distance between two points:

d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}

we have:

(-4, -6), (3, -7)

x_{1} = -4

y_{1} =-6

x_{2} = 3

y_{2} =-7

so we have:

d =\sqrt{(3-(-4))^{2}+ (-7-(-6))^{2}}\\\\d =\sqrt{(7)^{2}+ (-1)^{2}}\\\\d = \sqrt{49+ 1}\\\\d=\sqrt{50}\\\\d=5\sqrt{2}

4 0
3 years ago
Read 2 more answers
Help me PLEASE I need HELP
aivan3 [116]

Answer:

x=20

Step-by-step explanation:

180=2x+6x+20

180=8x+20

160=8x

20=x

5 0
3 years ago
The circular opening of a tunnel has a circumference of 36 meters. What is the diameter of the tunnel?
Elza [17]
C=2(3.14159...)R
You’re given the circumference, so all you need to do is plug that in and simplify.
36=2(3.14159...)R
36=6.28R
36/6.28=R
R is about 5.73
7 0
3 years ago
Jim Hunter has decided to retire to Florida in 10 years. What amount should Jim invest today so that he'll be able to withdraw $
ipn [44]
The following formula is applicable;
A=P(1+r)^n

Where,
A = Total amount accrued after 10 years (this is the amount from which the yearly withdrawals will be made from for the 30 years after retirement)
P=Amount invested today
r= Annual compound interest for the 10 years before retirement
n= Number of years the investments will be made.

Therefore,
A= Yearly withdrawals*30 years = $25,000*30 = $750,000
r= 9% = 0.09
n= 10 years

P= A/{(1+r)^n} = 750,000/{(1+0.09)^10} = $316,808.11
Therefore, he should invest $316,808.11 today.
7 0
4 years ago
Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?
Mademuasel [1]

Answer:- a.The given expression is equivalent to  \frac{x^{10}y^{14}}{729}



Given expression:- [\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}

=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}

=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}

=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}

=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}

=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}

Thus a. is the right answer.


6 0
3 years ago
Read 2 more answers
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