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3 years ago

Translate the following statement into an equation and solve for x

Mathematics

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

104 = -13 x -8

Step-by-step explanation:

The reason you got it wrong is, I'm assuming, because of the way you formatted it. Try this and see if it works.

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Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1, 3).

Romashka-Z-Leto [24]

The equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1, 3) in slope intercept form is $y = \frac{-1}{5}x + \frac{16}{5}$

<h3>Solution:</h3>

Given that a line is parallel to line x + 5y = 10 and passes through the point (1, 3)

We have to find the equation of line

The slope intercept form is given as:

y = mx + c -------- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Let us first find the slope of line

Given equation of line is x + 5y = 10

$5y = -x + 10\\\\y = \frac{-1}{5}x + \frac{10}{5}\\\\y = \frac{-1}{5}x + 2$

On comparing the above equation of line with slope intercept form,

$m = \frac{-1}{5}$

We know that slopes of parallel lines are equal

So the slope of line parallel to given line is also $m = \frac{-1}{5}$

Let us find the equation of line with slope m = -1/5 and passes through the point (1, 3)

$\text {substitute } m=\frac{-1}{5} \text { and }(x, y)=(1,3) \text { in eqn } 1$

$3 = \frac{-1}{5} \times 1 + c\\\\15 = -1 + 5c\\\\16 = 5c\\\\c = \frac{16}{5}$

Thus the required equation is:

$\text {substitute } m=\frac{-1}{5} \text { and } c=\frac{16}{5} \text { in eqn } 1$

$y = \frac{-1}{5}x + \frac{16}{5}$

Thus the required equation of line is found

3 0

3 years ago

Solve for x. Each figure is a trapezoid

Bingel [31]

Pretty sure x is 4 in #9

FG is congruent to ED, and GD and FE are parallel, so that means
16x + 1 = 65

Subtract 1 on both sides you get

16x = 64

Divide by 16 on both sides

x = 4

4 0

3 years ago

In a ∆ABC , angle A + Angle B = 125° and Angle B + Angle C = 150° . Find all the angles of ∆ABC.

mel-nik [20]

$\large\underline{\sf{Solution-}}$

Given that,

In triangle ABC

$\purple{\rm :\longmapsto\:\angle A + \angle B = 125 \degree \: - - - (1) }$

$\purple{\rm :\longmapsto\:\angle B + \angle C = 150 \degree \: - - - (2)}$

We know,

Sum of all interior angles of a triangle is supplementary.

$\purple{\rm :\longmapsto\:\angle A + \angle B + \angle C = 180\degree }$

On adding equation (1) and (2), we get 

$\purple{\rm :\longmapsto\:\angle A + \angle B + \angle B + \angle C = 125\degree + 150 \degree \:}$

$\purple{\rm :\longmapsto\:\angle A + \angle B + \angle C + \angle B = 275\degree \:}$

$\purple{\rm :\longmapsto\:180\degree + \angle B = 275\degree \:}$

$\purple{\rm :\longmapsto\:\angle B = 275\degree - 180\degree \:}$

$\purple{\rm :\longmapsto\:\angle B = 95\degree \:}$

On substituting the value in equation (1) and (2), we get

$\purple{\rm :\longmapsto\:\angle A + 95\degree = 125\degree }$

$\purple{\rm :\longmapsto\:\angle A = 125\degree - 95\degree }$

$\purple{\rm :\longmapsto\:\angle A = 30\degree }$

Also, from equation (2), we get

$\purple{\rm :\longmapsto\:95\degree + \angle C = 150\degree }$

$\purple{\rm :\longmapsto\:\angle C = 150\degree - 95\degree }$

$\purple{\rm :\longmapsto\:\angle C = 55\degree }$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{\angle A = 30\degree } \\ \\ &\sf{\angle B = 95\degree } \\ \\ &\sf{\angle C = 55\degree } \end{cases}\end{gathered}\end{gathered}$