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4 years ago

5-48÷4-4 What’s is the answer

Mathematics

2 answers:

Andrews [41]4 years ago

6 0

5-48 ÷ 4-4

-43 ÷ 0 = infinity

maksim [4K]4 years ago

5 0

Answer:-11

Step-by-step explanation:

following order of operations

48/4=12

5-12-4

-7-4=-11

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What is the Sector area with radius 6 and an angle of 50°? Use 3.14 for pi.<br>

nevsk [136]

Answer:

15.7 units²

Step-by-step explanation:

The sector area is 50/360 times the area of the circle (since the central area of a complete circle is 360° and that for the sector is 50°), or

Sector area = (50/360)(3.14)(6 units)² = 15.7 units²

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3 years ago

HELP PLS!! INCLUDE EXPLAINING THANK YOU

kap26 [50]

Answer:

Lucy paid the least amount of money and she paid $770

Step-by-step explanation:

The way to find dylans is

1000-80= $920

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3 years ago

WhT is 6811.09 rounded to the nearest ten

Ede4ka [16]

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The reason the 0 moved up to a 1 is because the number in the hundredth place was 5 or over. If it were 4 or under, then it would have remained a 0.

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3 years ago

What's 2 and 5/10 in standard decimal form

Evgesh-ka [11]

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3 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

3 years ago