PLease answer below easy question answer for 10 points and will give brainliest

Please answer number 7!!! ASAP!!!

Ilya [14]

Number 7 is Parallel!

7 0

3 years ago

Find an equation of a line passing through the point (8,9) and parallel to the line joining the points (2,7) and (1,5).

TiliK225 [7]

Answer:

2x - y - 7 = 0

Step-by-step explanation:

Since the slope of parallel line are same.

So, we can easily use formula,

y - y₁ = m ( x ₋ x₁)

where, (x₁, y₁) = (8, 9)

and m is a slope of line passing through (x₁, y₁).

and since the slope of parallel lines are same, so here we use slope of parallel line for calculation.

and, Slope = m = $\dfrac{y_{b}-y_{a}}{x_{b}-x_{a}}$

here, (xₐ, yₐ) = (2, 7)

and, $(y_{a},y_{b}) = (1, 5 )$

⇒ m = $\dfrac{5-7}{1-2}$

⇒ m = 2

Putting all values above formula. We get,

y - 9 = 2 ( x ₋ 8)

⇒ y - 9 = 2x - 16

⇒ 2x - y - 7 = 0

which is required equation.

3 0

3 years ago

Read 2 more answers

Which value of kkk makes 5-k+12=165−k+12=165, minus, k, plus, 12, equals, 16 a true statement?

trapecia [35]

Answer:

k=1

Step-by-step explanation:

Simplify 1/k

3 2 1

(((—+2)-(—•k))-4)-(—-2) = 0

k k k

2.1 Subtracting a whole from a fraction

2 2 • k

2 = — = —————

1 k

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

1 - (2 • k) 1 - 2k

——————————— = ——————

k k

Simplify 2/k

3 2 (1-2k)

(((—+2)-(—•k))-4)-—————— = 0

k k k

Simplify 3/k

3 2 (1-2k)

(((—+2)-(—•k))-4)-—————— = 0

k k k

5.1 Adding a whole to a fraction

3 (1 - 2k)

(((— + 2) - 2) - 4) - ———————— = 0

k k

6.1 Subtracting a whole from a fraction

2 2 • k

2 = — = —————

1 k

7.1 Subtracting a whole from a fraction

3 + 2 • k 2k + 3

————————— = ——————

k k

8.1 Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

2 2 • k

2 = — = —————

1 k

9.1 Pull out like factors :

2 - 2k = -2 • (k - 1)

10.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

-2•(k-1)

———————— • k = 0 • k

k

Now, on the left hand side, the k cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

-2 • (k-1) = 0

Equations which are never true:

10.2 Solve : -2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

10.3 Solve : k-1 = 0

Add 1 to both sides of the equation :

k = 1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

For 11 and 12, find... a.) The slope of the line parallel to the line that passes through the given points, and b.) The slope of

Minchanka [31]

(11)

we are given two points

(x1,y1)=(0,3)

(x2,y2)=(8,7)

so, we get

$x_1=0,y_1=3$

$x_2=8,y_2=7$

(a)

we know that slope of parallel lines are always equal

so, we can use formula

$m=\frac{y_2-y_1}{x_2-x_1}$

now, we can plug values

$m=\frac{7-3}{8-0}$

$m=\frac{4}{8}$

$m=\frac{1}{2}$ ............Answer

(b)

we know that slope of perpendicular line is -1/m

so,

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

now, we can plug value

$slope=\frac{-1}{\frac{1}{2}}$

$slope=-2$ ...............Answer

(12)

we are given two points

(x1,y1)=(5,-1)

(x2,y2)=(3,1)

so, we get

$x_1=5,y_1=-1$

$x_2=3,y_2=1$

(a)

we know that slope of parallel lines are always equal

so, we can use formula

$m=\frac{y_2-y_1}{x_2-x_1}$

now, we can plug values

$m=\frac{1+1}{3-5}$

$m=\frac{2}{-2}$

$m=-1$ ............Answer

(b)

we know that slope of perpendicular line is -1/m

so,

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

now, we can plug value

$slope=\frac{-1}{-1}$

$slope=1$ ...............Answer

7 0

3 years ago

2. Two angles have a common vertex but no common side. How can you tell whether the angles are vertical angles?

Maksim231197 [3]

Explanation:

The angles are <em>vertical angles</em> if the opposites of the rays forming one of the angles are the rays forming the other angle.

More formally, if V is the common vertex, and ...

R is a point on one of the rays forming Angle 1
S is a point on the ray that is the opposite of ray VR
T is a point on the other ray forming Angle 1
U is a point on the ray that is the opposite of ray VT

Then angle RVT and angle SVU are vertical angles.

___

Another way to say this is that points R, V, S are collinear, as are points T, V, U, and the two angles of interest are RVT and SVU.

If the above conditions cannot be met, then the angles are not vertical angles.

7 0

4 years ago