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3 years ago

What is anither way to write the equation 7/8x+3/4=-6

Mathematics

1 answer:

Andrei [34K]3 years ago

7 0

Answer:

7x + 6 = - 48

Step-by-step explanation:

Given

$\frac{7}{8}$ x + $\frac{3}{4}$ = - 6

Multiply through by 8 ( the LCM of 8 and 4 ) to clear the fractions

7x + 6 = - 48 ← same equation without fractions

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Mama L [17]

If x = the fuction hours
and y=kilometers walked
your equation should look like this
y=6.5(2.5)
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3 years ago

The table shows a direct variation.

marin [14]

The answer would be 4

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4 years ago

Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$

$\overrightarrow n=i(-9)-j(8)+k(2)$

$\overrightarrow n=-9i-8j+2k$

$\overrightarrow n=$

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0

3 years ago

Larry spends half of his workday teaching piano lessons. If he sees 6 students, each for the same amount of time, what fraction

Marina CMI [18]

Answer:

1/12

Step-by-step explanation:

bc ik

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3 years ago

Solve the following system by any method -8x-8y=0 -8x+2y=-20

iVinArrow [24]

$\fbox{Solution by using Matrix}$

$\text{Linear Equation} = -8x-8y=0 , -8x+2y=-20$

$\text{Rewrite the linear equations above as a matrix}$

$\left[\begin{array}{ccc}8&-8&0\\-8&2&-20\\\end{array}\right]$

$\text{Apply to Row2 : Row2 + Row1}$

$\left[\begin{array}{ccc}8&-8&0\\-8&-6&-20\\\end{array}\right]$

$\text{ Simplify rows}$

$\left[\begin{array}{ccc}8&-8&0\\0&1&10/3\\\end{array}\right]$

$\text{Note: The matrix is now in echelon form.}$ $\text{The steps below are for back substitution.}$

$\text{Apply to Row1 : Row1 + 8 Row2}$

$\left[\begin{array}{ccc}8&0&80/30\\0&1&10/3\\\end{array}\right]$

$\text{ Simplify rows}$

$\left[\begin{array}{ccc}1&0&10/3\\0&1&10/3\\\end{array}\right]$

$\text{Therefore, the solution is}$

$x= \dfrac{10}{3} \ \text{and} \ \ y=\dfrac{10}{3}$

7 0

3 years ago