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3 years ago

13

Of the first 100 customers to walk into a grocery store, 9 bought flowers. What is that percent of the customers that bought flo

wers?

2 answers:

klasskru [66]3 years ago

7 0

9% of people because 9 of them brought flowers, and 9/100 is equal to 0.09.

Naya [18.7K]3 years ago

4 0

percent means out of 100

9/100 = 9 percent

9% bought flowers

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Can you guys help please?

Vadim26 [7]

Answer:

25%

Step-by-step explanation:

There are 3 shaded and there are 12 in all, 3 is 25% of 12

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3 years ago

Tell whether the points appear to represent a linear function, an exponential function, or neither? Explain please.

mario62 [17]

Answer:

Exponential

Step-by-step explanation:

Linear is easy to remember because it goes in a straight LINE.

Only other option is a quadratic and this does not create a parabola

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3 years ago

Find the circumference and area of a circle with diameter 22 ft. Express your answer in terms of π.

Ray Of Light [21]

Answer:

69.12ft

Step-by-step explanation:

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3 years ago

Select the correct answer from each drop-down menu.

Elodia [21]

Answer:

The left end approaches to + Infinite being an exponential function, and the right end to 0

Step-by-step explanation:

we have to remember the exponential function, the best way to find the answer is plotting the graph or arranging a table of values.

As you can see in the attached graph the y axis gets closer and closer to 0 as it moves forward in the x axis, and as it moves to the left the y axis starts increasing rapidly.

Also you got to keep in mind the way that functions behave in terms of the sign of its variable. for example the 10 in this equation only makes the curve to get wider, but if you change the sign to minus, the answer would be different.

8 0

4 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago