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Lady bird [3.3K]

3 years ago

8

An object is translated by (x - 2, y - 6). If one point in the pre-image has the coordinates (-3, 7), what would be the coordina

tes of its image?
PICTURE INCLUDED

2 answers:

Rudik [331]3 years ago

5 0

Answer:

-5,1

Step-by-step explanation:

IRINA_888 [86]3 years ago

4 0

Answer:

-5,1

Step-by-step explanation:

i hope this helps and i hope it isnt wrong

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A quadratic function of the form 0=ax2+bx2+c has a discriminant value of 0 how many real number solutions does the equation have

xxMikexx [17]

The discriminant is b²-4ac
when the discriminant is 0, there is only one solution.

4 0

3 years ago

Write an equation that represents a relationship that is not<br> proportional when m= 3 and b= -6.

Rainbow [258]

Answer:

y=3x+-6

Step-by-step explanation:

plug it into y=mx+b form

hope this helps

mark brainiest pls

4 0

2 years ago

What is 96 written in expanded form?

natita [175]

Answer:

90 + 6

Step-by-step explanation:

90 + 6 = 96

<u>ALTERNATIVE</u><u>:</u>

96 - 6 = 90

6 0

2 years ago

URGENT HELP NEED!!! I WILL REWARD LOTS OF POINTS FOR THIS QUESTION!!! ANY IRRALEVENT OR SILLY QUESTION WILL BE INSTANTLY REPORTE

Ipatiy [6.2K]

Let the points be A,B,C

A(0,0)
B(24,16)
C(8,20)

#A

$\\ \bull\sf\leadsto AB=\sqrt{(24-0)^2+(16-0)^2}=\sqrt{24^2-16^2}=\sqrt{576-256}=\sqrt{320}=17.4units$ (leg3)

$\\ \bull\sf\leadsto AC=\sqrt{(8-0)^2+(20-0)^2}=\sqrt{8^2+20^2}=\sqrt{64+400}=\sqrt{464}=21.4units$ (leg1)

$\\ \bull\sf\leadsto BC=\sqrt{(24-8)^2+(26-20)^2}=\sqrt{16^2+(-4)^2}=\sqrt{256+16}=\sqrt{272}=16.4units$ (leg2)

#B

We have to find perimeter

$\\ \bull\sf\leadsto Perimeter=AB+AC+BC$

$\\ \bull\sf\leadsto Periemter=17.4+16.4+21.4$

$\\ \bull\sf\leadsto Perimeter=55.2units$

#C

The length of leg3=17.4units

5 0

3 years ago

Read 2 more answers

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago