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Anna [14]
3 years ago
10

The dashed triangle is the image of the solid triangle. The center of dilation is (−5, 7) .

Mathematics
2 answers:
Zinaida [17]3 years ago
6 0

Answer:

Just took the test and the answer is -0.25

Hope this helps future people :-)

adell [148]3 years ago
3 0
Before dilation:
the height of the triangle is h₁ = 7 - (-1) = 8
the width of the triangle is w₁ = 7 - (-5) = 12

After dilation:
the height of the is h₂ = 9-7 = 2
the width of the triangle is w₂ = -5 - (-8) = 3

The scale factor is h₂/h₁ = 2/8 = 1/4 = 0.25
Also, the scale factor is w₂/w₁ = 3/12 = 1/4 = 0.25

Answer: The scale factor is 0.25

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Step-by-step explanation:

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4 0
3 years ago
Help me please i will mark brainliest
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Step-by-step explanation:

The answer is 1/5

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In ΔABC, which trigonometric ratio equals 32?
Nezavi [6.7K]
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tan C = sin C / cos C

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cos C = 2 / 3.61

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6 0
3 years ago
A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
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