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klemol [59]
3 years ago
10

Find the function rule for g(x). The function f(x) = x2 . The graph of g(x) is f(x) translated to the right 8 units and down 7 u

nits. What is the function rule for g(x)?
A. g(x) = (x + 7)2 + 8
B. g(x) = (x + 8)2 + 7
C. g(x) = (x - 7)2 - 8
D. g(x) = (x - 8)2 - 7
Mathematics
1 answer:
anygoal [31]3 years ago
4 0
F(x)=x^2
Translated to the right 8 units: h(x)=(x-8)^2
and down 7 units: g(x)=(x-8)^2-7

Answer: Option D. g(x)=(x-8)^2-7
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If the length of the minute hand of the clock in London commonly known as “Big Ben” is 11.25’ from the center of the clock to th
rodikova [14]
First, we should figure out the area of the entire face of the clock because we need that information to solve the problem. The formula for the area of a circle is A=pi*r^2. Since we know that r (radius) is equal to 11.25 feet, we can plug this in for r and solve for A: A=pi*11.25^2 which equals A=397.61 ft^2 rounded to the nearest hundredth. 

Now, to find the area the hand sweeps over in 5 minutes, we should determine how much of the clock the hand sweeps over in 5 minutes. Think about it like this: since 5 minutes goes into 60 minutes 12 times (60/5=12), then 5 minutes is one twelfth of the clock's face. Therefore, we are going to divide the total area by 12 (397.61/12) to get 33.13 ft^2, so the answer is C.

I hope this helps.
5 0
3 years ago
Joanne was hired by a florist to make large boes for holidays wreaths each roll of ribbon has 20 yards joanne used 3 1/2 rolls o
alukav5142 [94]

Answer: Each bow needs 4.67 yards of ribbon.

Step-by-step explanation:

We know that each roll has 20 yd.

She used 3 + 1/2 rolls, then the total length used will be (3 + 1/2) times the length of each roll, this is:

(3 + 1/2)*20 yd = (6/2 + 1/2)*20yd = (7/2)*20yd = 70 yd.

With tose 70 yards of ribbon, she made 15 bows, then the length of ribbon used in each bow will be equal to the quotient between the total length used and the number of ribbons made with it, this is:

70yd/15 = 4.67 yards

Each bow needs 4.67 yards of ribbon.

6 0
2 years ago
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
Write a rule for each pattern. Then use your rule to find the next two terms in the pattern. The numbers are: 1, 4, 9, 16, 25, 3
Anika [276]
When n is the nth term in the sequence, n squared is the rule.

The next two terms are 64, 81

Good luck!
3 0
3 years ago
Read 2 more answers
A cone has a circular base with a diameter of 18 inches. The height of the cone is 40 inches
OLEGan [10]
U have given half of the question, I mean, do we have to find something or what? please mention :)
7 0
3 years ago
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