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AleksandrR [38]
3 years ago
13

A new doughnut shop lost $624 during its first 13 days of business. If it lost the same amount of money each day, what was the l

oss each day?
Which choice shows the correct quotient and answer?

A. (−624)÷13=−48; The doughnut shop lost $48 each day.

B. (−624)÷13=48; The doughnut shop made a profit of $48 each day.

C . (−624)÷13=−611; The doughnut shop lost $611 each day.

D (−624)÷13=8,112; The doughnut shop made a profit of $8,112 each day.
Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

A. (−624)÷13=−48; The doughnut shop lost $48 each day.

Step-by-step explanation:

Because the donut shop lost $624 in the first 13 days the way to find the amount they lost per day is with division meaning that we would ualf to divided 624 divided by 13 first and the answer to this equation is 48 and to make it correct we have to add on a the negative sign to the 48 since they were losing $48 a day. Therefore,the doughnut shop lost $48 per day and the answer is A.

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Answer:

4!

Step-by-step explanation:

In word MAXIMUM, consonants are M & X. According to the question itself, consonants should not occur together.

So, we have to put vowels at each even position; only then the condition will be satisfied.

We can put this in 3! ways (permutations of A, I & U).  

For odd positions we can put 3-M's & a X. Total ways is 4!/3!   (as repetitions of M).

Hence, the answer is 4!/3! x 3! = 24 ways.  

24=4!.

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3 years ago
hich of the following expressions is equivalent to |x + 4| < 5? A. –5 > x + 4 < 5 B. –5 < x + 4 < 5 C. x + 4 <
Airida [17]

Answer:

option B

Given : |x + 4| < 5

A. –5 > x + 4 < 5

B. –5 < x + 4 < 5

C. x + 4 < 5 and x + 4 < –5

D. x + 4 < 5 or x + 4 < –5

In general , |x|< n where n is positive

Then we translate to -n < x < n

|x + 4| < 5

5 is positive, so we translate the given absolute inequality to

-5 < x+4 < 5

So option B is correct

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3 years ago
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omeli [17]

Answer:

c

Step-by-step explanation:

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3 years ago
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Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
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so the particular solution to the IVP is

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4 0
3 years ago
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inysia [295]

Answer: P= QRS

#edmentumlivesmatter

Step-by-step explanation:

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3 years ago
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