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3 years ago

Solve for v. -8 +5v=-33

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

Answer:

Solve

-8 +5v=-33

So 5v=-33+8

(-8 became 8 because we brought it back to the other side to have the v 《5v》 alone and the numbers alone )

5v= -25

So v=-25/5

We simplify -25/5

v=-5

Dima020 [189]3 years ago

3 0

Answer:

v = -5

Step-by-step explanation:

Step 1: Write out equation

-8 + 5v = -33

Step 2: Add 8 on both sides

-8 + 8 + 5v = -33 + 8

5v = -25

Step 3: Divide both sides by 5

5v/5 = -25/5

v = -5

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3 years ago

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The difference between the square roots of a number is 30. what is the answer?

Sveta_85 [38]

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3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

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4 years ago

If b=4 units and h=5 units, the area of the triangle is _____<br> units2.

mestny [16]

Answer:

Step-by-step explanation:

Area = 1/2 b h

= 1/2 * 4 * 5

= 10 unit^2.

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2 years ago

Solve for v. -8 +5v=-33​

Solve for v. -8 +5v=-33