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alexandr1967 [171]

3 years ago

14

The machine bottled 8,000 bottles in 4 hours. if the rate of the machine were doubled, how long would it take the machine to bot

tle 40,000 bottles?

1 answer:

valina [46]3 years ago

4 0

It would take 20 hours since 8,000 times 5 is 40,000 so 4 times 5 is 20.

Logic.

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John borrowed a certain amount of money from Tebogo at a simple interest rate of 8,7% per year.

Anestetic [448]

Answer:

Step-by-step explanation:

5 0

2 years ago

14 by 15 feet garden. if one bag of soil covers 20 square feet how many bags of soil

lesantik [10]

14*15 is 210 which his the entire area. now divide 210/20 you get 10.5 so you will need at least 11 bags

6 0

3 years ago

Read 2 more answers

A moving truck is rented at $1,000 per day plus a charge per hour of use. The moving truck was used for 9 hours one day, and the

Advocard [28]

1000 is our constant because the truck was only used one day. We also know they charge a certain amount per hour, but we do not know how much. In this case they charged that amount 9 times because the truck was used for 9 hours. The total of the constant plus the 9hour fee is 2700 dollars:

1000(1) + 9x = 2700

The reason we multiply 1000 by 1, is because we only used the truck for one day.

Solve for x

9x = 2700 - 1000 = 1700

X = 1700/9 = 188.88

The hourly fee is $188.88
The question is asking for an equation for hours of use though, so the answer is

1000(1) + 9(188.88) = C

C represents cost or charge.

4 0

3 years ago

Read 2 more answers

PLS ANSWER ASAP THANKS

Afina-wow [57]

Answer:

x=54°

Step-by-step explanation:

x+36°=90°

x=90°-36°

x=54°

<em><u>HOPE IT HELPS U </u></em>

<em><u>HAVE A GREAT DAY</u></em>

3 0

2 years ago

The matrix A= (−3 0 1, 2 −4 2, −3 −2 1) has one real eigenvalue. Find this eigenvalue, its multiplicity, and the dimension of th

Aliun [14]

Answer:

a) -4

b) 1

c) 1

Step-by-step explanation:

a) The matrix A is given by:

$A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right]$

to find the eigenvalues of the matrix you use the following:

$det(A-\lambda I)=0$

where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:

$A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right]$

and by taking the determinant:

$[(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0$

and the roots of this polynomial is:

$\lambda_1=-4\\\\\lambda_2=-1+i\sqrt{3}\\\\\lambda_3=-1-i\sqrt{3}$

hence, the real eigenvalue of the matrix A is -4.

b) The multiplicity of the eigenvalue is 1.

c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)

3 0

2 years ago