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2 years ago

Find the vertices of the image after the given transformation (x,y)—> (2x, 1/3y)

Mathematics

1 answer:

Olin [163]2 years ago

8 0

Answer:

see explanation

Step-by-step explanation:

the translation rule (x, y ) → (2x, $\frac{1}{3}$ y )

means multiply the x- coordinate by 2 and multiply the y- coordinate by $\frac{1}{3}$ ), so

A (- 1, - 3 ) → A' (- 2, - 1 )

B (4, - 3 ) → B' (8, - 1 )

C (4, 3 ) → C' (8, 1 )

D (- 1, 3 ) → D' (- 2, 1 )

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Find the value of x in each case:

kicyunya [14]

I think the value of x is 15

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3 years ago

Read 2 more answers

F(x)=3×^2+5×+4×and g(×)=3×find (f/g)(×)<br>

serious [3.7K]

Answer:

Step-by-step explanation:

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3 years ago

What is the contrapositive of the statement below?

algol13

Answer:

B. -y = -x

Step-by-step explanation:

When the values on both sides of the equal sign are changed, it becomes a negative number.

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2 years ago

Solve the equation and check the solution4.4x-4=3.4x

PolarNik [594]

$4.4x-4=3.4x$

Substract '-3.4x' at LHS nad the RHS of the above expression.

$\begin{gathered} 4.4x-4-3.4x=3.4x-3.4x \\ 4.4x-3.4x-4=0 \\ x-4=0 \end{gathered}$

Add '4' on both LHS and RHs of the above expression.

$undefined$

4 0

1 year ago

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

8 0

1 year ago