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Sonbull [250]
3 years ago
13

There was no snow on the ground when it started falling at midnight at a constant rate of 1.5 inches per hour. At

Mathematics
1 answer:
anyanavicka [17]3 years ago
3 0

Answer:

The depth of snow at 9:00 am was 19 inches.

Step-by-step explanation:

Consider the provided information.

Let x represents the number of hours and S(x) represents the depth of snow.

There was no snow on the ground when it started falling at midnight at a constant rate of 1.5 inches per.

That means the depth of snow will be:

S(x)=1.5x\ \ \ 0\leq x< 4

At  4:00 a.m., it starting falling at a constant rate of 3 inches per hour,

S(x)=3(x-4)+6\ \ \ 4\leq x

7:00 a.m. to 9:00 a.m., snow was  falling at a constant rate of 2 inches per hour.

S(x)=2(x-7)+15\ \ \ 7\leq x\leq 9

The required piece-wise linear function is

\left\{\begin{matrix}1.5x & 0\leq x

We need to find the how deep was the snow at 9:00 am

For 9:00 am we will choose the function which can take the value of x=9

Substitute the value of x=9 in S(x)=2(x-7)+15

S(x)=2(9-7)+15

S(x)=2(2)+15

S(x)=19

Hence, the depth of snow at 9:00 am was 19 inches.

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There is 3.28084 feet in a meter
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Jack invested $2,900 in an account paying an interest rate of 8 % compounded
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Answer: $628

Step-by-step explanation:

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3 years ago
You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o
Aleksandr [31]

Answer:

Therefore the circumference of the circle is =\frac{20\pi}{4+\pi}

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

\Rightarrow s=\frac{10-\pi r}{2}

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

A=\pi r^2+ (\frac{10-\pi r}{2})^2

\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2

\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}

\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25

For maximum or minimum \frac{dA}{dr}=0

Differentiating with respect to r

\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi

Again differentiating with respect to r

\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}    > 0

For maximum or minimum

\frac{dA}{dr}=0

\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0

\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}

\Rightarrow r=\frac{10}{4+\pi}

\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0

Therefore at r=\frac{10}{4+\pi}  , A is minimum.

Therefore the circumference of the circle is

=2 \pi \frac{10}{4+\pi}

=\frac{20\pi}{4+\pi}

4 0
3 years ago
Please help mee l don’t understand new topic please
jasenka [17]

Answer to first four-

first one- 2f+2f=60

4f=60

f=15

second one-18+2d=38

2d=20

d=10

(short side is 9)

third one-2a+a+5+a+5=26

4a+10=26

a=4

long side is 9

short side is 8

fourth one-3xy+3xy+xy=63

7xy=63

xy=9

long side is 27

4 0
2 years ago
1. 4x + 3(x-9) = 19
Dmitry_Shevchenko [17]

1) 4x + 3(x-9) = 19

or, 4x + 3x - 27 = 19

or, 7x = 19+27

or, 7x = 46

or, x = 46/7

◆ x = 6.57

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

or, 2x - 8 + 5x = 3

or, 7x = 3+8

or, x = 11/7

◆ x = 1.57

3) 60 = 8y - 3(y-10)

or, 60 = 8y - 3y + 30

or, 60 - 30 = 5y

or, y = 30/5

◆ y = 6

4) 40 + 3c = 3 + 9(c-2)

or, 40 + 3c = 3 + 9c - 18

or, 40 + 3c = 9c - 15

or, 9c - 3c = 40 + 15

or, 6c = 55

◆ c = 9.166

I hope you understand..

Thank you...♥♥

5 0
2 years ago
Read 2 more answers
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