8(t+2) - 3(t-4) = 6(t-7) + 8

Mathematics

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

t = 50 is the solution for the given expression 8(t+2) - 3(t-4) = 6(t-7) + 8

Step-by-step explanation:

Here, the given expression is 8(t+2) - 3(t-4) = 6(t-7) + 8

Solving the given expression, we get

8t + 16 -3t + 12 = 6t - 42 + 8

or, 5t + 18 = 6t - 34

⇒ 5t - 6t = -34 -16

or , -t = 50

or, t = 50

⇒ t = 50 is the solution for the given expression 8(t+2) - 3(t-4) = 6(t-7) + 8

You might be interested in

write an equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

dimaraw [331]

The equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is $y - 3 = \frac{-7x}{2}+ \frac{21}{4}$

<h3>Solution:</h3>

Given that we have to write equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

Let us first find the slope of given line AB

The slope "m" of the line is given as:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Here the given points are A(-2,2) and B(5,4)

$\text {Here } x_{1}=-2 ; y_{1}=2 ; x_{2}=5 ; y_{2}=4$

$m=\frac{4-2}{5-(-2)}=\frac{2}{7}$

Thus the slope of line with given points is $\frac{2}{7}$

We know that product of slopes of given line and slope of line perpendicular to given line is always -1

$\begin{array}{l}{\text {slope of given line } \times \text { slope of perpendicular bisector }=-1} \\\\ {\frac{2}{7} \times \text { slope of perpendicular bisector }=-1} \\ \\{\text {slope of perpendicular bisector }=\frac{-7}{2}}\end{array}$