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ra1l [238]
2 years ago
11

Mrs. Hidalgo paid $30 for 4 students to visit an art museum. Find the cost for 20 students.

Mathematics
1 answer:
Bond [772]2 years ago
8 0

Answer:

The cost for 20 students would be $120

Step-by-step explanation:

We know that 4x5=20 therefore we will multiply 30x4 which will give us 120.

Our final answer would be $120

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Step-by-step explanation:

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Step-by-step explanation:

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Please help! functions operations. explain please
ValentinkaMS [17]

Answer:

a. 40

Step-by-step explanation:

Given

g(x)=2x and h(x)=x^2+4

(h\circ g)(x)=h(g(x))

(h\circ g)(x)=h(2x)

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(h\circ g)(x)=(2x)^2+4

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7 0
3 years ago
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Clara found the product of 3 – 6y2 and y2 + 2. Her work is shown below. (3 – 6y2)(y2 + 2) = 3(y2) + (–6y2)(2) = 3y2 – 12y2 = –9y
vladimir2022 [97]

Answer:

No, She did not use the distributive property correctly.

Step-by-step explanation:

While multiplying (3-6y^2)(y^2+2), he did not use distributive property correctly. She only multiplied 3 with y^2 and not with 2. Also she multiplied -6y^2only with 2 and not with y^2.

6 0
2 years ago
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
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