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klemol [59]
3 years ago
14

A store is having a sale with a 19% discount on all merchandise. Write an

Mathematics
1 answer:
inna [77]3 years ago
7 0

Answer:

3

Step-by-step explanation:

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Let A={1, 2, 3, 4} and B= {2,5,7} find the set A U B
vaieri [72.5K]
A ∪ B is {1, 2, 3, 4, 5, 7}

Set up the union notation of set {1,2,3,4} and {2,5,7}.
{1,2,3,4}∪{2,5,7}

The union of two sets is the set of elements which are in either sets.
{1,2,3,4,5,7}
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3 years ago
Use the image below to answer the following parts.
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Answer:

Step-by-step explanation:

Part A

Since, the given angles are formed at a point are on a straight line, sum of these angles will be 180°.

Part B

(3x - 5)° + (4x + 2)° + (2x + 3)° = 180°

9x = 180

x = 20

Part C

m∠ADB = (3x -  5)°

             = (3×20 - 5)°

             = 55°

m∠DBK = (4x + 2)°

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m∠KBC = (2x + 3)°

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6 0
3 years ago
What is 10 to the 8th power multiplied by 30 subtract 3
IceJOKER [234]

Answer:

21

Step-by-step explanation:

4 0
3 years ago
Please help i dont want to fail..
Margarita [4]

Answer: the answer is d pls give brainlEst

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Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

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

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(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

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
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