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3 years ago

What is the image of the point (5,-3) after a reflection over the line y = x?

Mathematics

2 answers:

timofeeve [1]3 years ago

6 0

Answer:

(-5,3)

Step-by-step explanation:

Rufina [12.5K]3 years ago

6 0

Answer:

it's C.

Hope this helps!!

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Explain two different ways that you can solve the exponential equation 3^x+2=8 .

Nonamiya [84]

Answer:

solving a quadratic equation by factoring

using the quadratic formula

Step-by-step explanation:

3 0

3 years ago

Brandee makes an hourly wage. In the last pay period, she earned $800 for regular hours and $240 for overtime hours. Her overtim

Aleonysh [2.5K]

Step-by-step explanation:

Overtime rate= r+50%= 1.5r

Regular hours= 800/r

Overtime hours= 240/1.5r

Total hours worked

h=800/r+240/1.5r

h= 800/r+160/r

h=960/r

r=960/h

3 0

4 years ago

Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the ot

Softa [21]

Answer:

a. Mean = 4.5, Standard Deviation = 1.775

b. 0.0152

Step-by-step explanation:

Given

n = 15 purchasers

p = Success = 30%

p = 0.3

q =Failure = 70%

q = 0.7

Mean = np

Mean = 15 * 0.3

Mean = 4.5

Standard Deviation = √Variance

Variance = npq

Variance = 15 * 0.3 * 0.7

Variance = 3.15

Standard Deviation = √3.15

Standard Deviation = 1.774823934929884

Standard Deviation = 1.775 ---------- Approximated

The probability that the number who want new copies is more than two standard deviations away from the mean value

Standard Deviation = 1.775

Mean = 4.5

2 Standard Deviation and Mean = 2 * 1.775 + 4.5

= 3.55 + 4.5

= 8.05

P(X>8.05) = P(9) + P(10) +........+ P(15)

Using the binomial distribution

(p + q) ^ n where p = 0.3, q = 0.7 , n = 15

Expanding (p+q)^n where n = 15 and r > 8

We have

15C9 p^9 q^6 + 15C10 p^10 q^5 + 15C11 p^11 q⁴ + 15C12 p^12 q³ + 15C13 p^13 q² + 15C14 p^14 q + p^15

= 5005 (0.3)^9 (0.7)^6 + 3003 (0.3)^10 (0.7)^5 + 1365 (0.3)^11 (0.7)⁴ + 455 (0.3)^12 (0.7)³ + 105 (0.3)^14 (0.7)² + 15 (0.3)^14 (0.7) + 0.3^15

=0.015234

7 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

What is the area of a circular fountain that has a 6 ft. diameter?

Dominik [7]

Answer: B. 28.26 ft2

Step-by-step explanation: I did this in school and got it right.

3 0

2 years ago

Read 2 more answers