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frozen [14]
3 years ago
6

Angela is always forming study groups before tests. What type learning preference might she have? A. Visual c. Auditory b. Inter

personal d. Intrapersonal Please select the best answer from the choices provided
Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

Answer:

Learning preference that Angela prefer is interpersonal.

Step-by-step explanation:

Learning preference that Angela prefer is interpersonal. As it is defined as the interaction which done within one close group where member interact and share ideas among each other.

while intrapersonal refer to the oneself communications, visual and auditory communication used some video lesson and audio guide respectively to enhance one knowledge

Vladimir [108]3 years ago
6 0

Answer:

B Hope it helps

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Nola purchased two. 5 pounds of cheese for 10. $50 her mother purchased 3 pounds of the same cheese for $12.60 the cost of the c
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Answer: 1 pound of cheese cost $4.2

Step-by-step explanation:

Nola purchased 2.5 pounds of cheese for 10. $50

Her mother purchased 3 pounds of the same cheese for $12.60

If the cost of the cheese varies directly with the number of pounds purchased, it means that the higher the pounds of cheese purchased, the higher the cost

To determine the the cost of 1 pound of cheese, we wound divide the the cheese by the number of pounds bought.

It becomes 10.5/2.5 = 4.2

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Consider the function f(x)= x(x-4)
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4 years ago
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beth made a batch of blueberry and banana muffins. Each batch is 6 muffins. She makes 2.5 batches of blueberry muffins. How many
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The answer should be 7.5 batches of banana. I hope this helps
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Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

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

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
The product of a number and 5 is increased by 12 the result is 32 find the number​
frozen [14]

Answer:

4

Step-by-step explanation:

5 x 4 = 20

+ 12 = 32

3 0
3 years ago
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