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3 years ago

-12c = -780. How do I solve that?

Mathematics

1 answer:

Alex Ar [27]3 years ago

5 0

Here’s your answer I hope this helps

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Enlarge the triangle by scale factor of -1/3

Bess [88]

The attached image represents the triangle after it has been enlarged

<h3>How to enlarge the triangle?</h3>

The given parameters are:

Scale factor, k = -1/3
Center, (a,b) = (-1,2)

From the graph, the coordinates of the triangle are:

A = (-4,2)

B = (-4,-4)

C = (-1,-4)

The rule of dilation is calculated using:

$(x,y) \to (k(x - a) + a, k(y - b) + b)$

So, we have:

$A' = (-\frac 13(-4 +1) - 1, -\frac 13(2 - 2) + 2)$

A' = (0, 2)

$B' = (-\frac 13(-4 +1) - 1, -\frac 13(-4 - 2) + 2)$

B' = (0, 4)

$C' = (-\frac 13(-1 +1) - 1, -\frac 13(-4 - 2) + 2)$

C' = (-1, 4)

See attachment for the image of the triangle, after it has been enlarged

Read more about dilation at:

brainly.com/question/3457976

#SPJ1

4 0

3 years ago

Use the chart below to answer the questions:

MrRissso [65]

Answer:

Good question IM gonna come back and awnser this question dont worry this wont be the last of Me

Step-by-step explanation:

6 0

3 years ago

Determine the equivalent system for the given system of equations.<br> 4x - 5y = 2<br> 3x - y = 8

kondor19780726 [428]

4x − 5y = 2
7x + 6y = 10

4 0

3 years ago

Find the value of x which ABCD must be a parallellogram?<br> x =

svetoff [14.1K]

$ABCD\ is\ parallelogram\ then\ :\\\\\underbrace{15x+2=9x+20}_{alternate\ angles}\ \ \ |subtract\ 2\ from\ both\ sides\\\\15x=9x+18\ \ \ \ |subtract\ 9x\ from\ both\ sides\\\\6x=18\ \ \ \ |divide\ both\ sides\ by\ 6\\\\\boxed{x=3}$

3 0

3 years ago

Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k

Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

7 0

3 years ago