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3 years ago

M(4, 2) is the midpoint of . The coordinates of S are (6, 1). What are the coordinates of R? (8, 0) (2, 3) (1.5, 3) (5, 1.5)

1 answer:

6 0

Assuming M is midpoint of RS

the midpoint formula is

x term of the midpoint is average of the x value of the 2 points
y term of the midpoint is average of the y value of the 2 points

if R is (x,y)

S is (6,1)

x value of midpoint is 4
x value of S is 6
x value os R is x

so
average of x and 6 is 4,
(x+6)/2=4,
x+6=8,
x=2

y value of midoint is 2
y value of S is 1
y value of R is y

average of 1 and y is 2
(y+1)/2=2
y+1=4
y=3

the midpoint is (2,3)

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Factor 9x^2+66xy+121y^2

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Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

[ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.

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3 years ago

The water tank in your school holds 20 liters of water. One day it was 34 full. That day, 14 liters of tank capacity was used up

Alex787 [66]

Answer:

Step-by-step explanation:

The water tank in the school hold 20 liters of water

One day it was 34 liters full

14 liters of water was used up

Therefore the quantity of water left can be calculated as follows

= 34-14

= 20

Hence 20 liters of water remains in the tank

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3 years ago

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3 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago

Tereka and Emma each deposit money in savings accounts. The amount of money, in dollars, in Tereka’s account after x years is mo

Over [174]

Given:

Amount of money in Tereka’s account : $T(x)=472(1.04)^x$

Amount of money in Emma’s account : $E(x)=485(1.09)^x$

To find:

The difference between initial amounts Tereka and Emma deposit.

Solution:

Substitute x=0 in the given functions, to get the initial amount of money they deposit.

$T(0)=472(1.04)^0$

$T(0)=472(1)$

$T(0)=472$

So, the initial amount Tereka deposit is $472.

$E(0)=485(1.09)^0$

$E(0)=485(1)$

$E(0)=485$

So, the initial amount Emma deposit is $485.

Difference between the initial amounts Tereka and Emma deposit is

$485-472=13$

Therefore, the difference between initial amounts Tereka and Emma deposit is $13. Emma deposit $13 more than Tereka.

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3 years ago