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2 years ago

A) Round 8.6078 to 2 decimal places

Mathematics

2 answers:

Lady bird [3.3K]2 years ago

8 0

Answer: 8.61

Step-by-step explanation: Since the third decimal (7) is greater than 5, the second decimal is rounded up. Therefore, 8.6078 rounded to two decimal places is 8.61.

Vlada [557]2 years ago

5 0

Answer:

8.6

Step-by-step explanation:

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ch4aika [34]

Answer:

43.25-7-(t*1.15)

Step-by-step explanation:

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2 years ago

Read 2 more answers

The density of a certain material is such that it weighs 6 pounds per pint of volume.

ella [17]

Answer:

Density = 5751.66 gram per litre

Step-by-step explanation:

It is given that the density of a certain material 6 pounds per pint.

The conversion from pint to liter is as follows :

1 pint = 0.473 litre

The conversion from pounds to kg is as follows :

1 pound = 453.59 kg

We need to express the density from pounds per pint to grams per liter.

1 pound per pint = 958.611 gram per litre

So,

6 pound per pint = 958.611 × 6 gram per litre

6 pound per pint = 5751.66 gram per litre

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2 years ago

A system of equations is shown below,

Vikki [24]

Answer: No solution
Step by step:
10x-2(5x+2)=12
10x-10x-4=12
0x=16
No solution

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2 years ago

The sum of three even consecutive integers if the first integer is y.

user100 [1]

Answer:

y + 1 + y + 2 + y + 3 =3y+6

Step-by-step explanation:

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2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

2 years ago