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3 years ago

11b+3(b+1)=14b+3

Mathematics

2 answers:

Arte-miy333 [17]3 years ago

5 0

Answer:

(

)

11b+3(b+1)=14b+3

Solve

Distribute

(

)

Combine like terms

Subtract

from both sides of the equation

Simplify

Subtract

14b

from both sides of the equation

Simplify

koban [17]3 years ago

4 0

Answer:

The equation is an identity: it is true for all values

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Answer:

9 gems

Step-by-step explanation:

$2 + 2 + 5 = 9$

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Which line is parallel to a line that has a slope of 3 and a y-intercept at (0, 0)?

anyanavicka [17]

we know that

If two lines are parallel , then their slopes are the same

we will proceed to calculate the slope of each line to determine the solution.

The formula to calculate the slope m between two points of the line is equal to

$m=\frac{(y2-y1)}{(x2-x1)}$

case N 1) line AB

Let

$A(-4,3)\\B(4,3)$

substitute the values

$m=\frac{(3-3)}{(4+4)}$

$m=\frac{(0)}{(8)}$

$m=0$

$0\neq 3$

therefore

The line AB is not the solution

vase N 2) line FG

Let

$F(-3,-1)\\G(3,-3)$

substitute the values

$m=\frac{(-3+1)}{(3+3)}$

$m=\frac{(-2)}{(6)}$

$m=-1/3$

$-1/3\neq 3$

therefore

The line EG is not the solution

case N 3) line CD

Let

$C(-3,0)\\D(3,2)$

substitute the values

$m=\frac{(2-0)}{(3+3)}$

$m=\frac{(2)}{(6)}$

$m=1/3$

$1/3\neq 3$

therefore

The line CD is not the solution

case N 4) line HJ

Let

$H(-1,-4)\\J(1,2)$

substitute the values

$m=\frac{(2+4)}{(1+1)}$

$m=\frac{(6)}{(2)}$

$m=3$

$3=3$

therefore

The line HJ is the solution

therefore

the answer is the option

line HJ

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Substitute x = -5
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3 years ago

HELP ASAP (Geometry)

Andrei [34K]

1) Parallel line: $y=-2x-3$

2) Rectangle

3) Perpendicular line: $y = 0.5x + 2.5$

4) x-coordinate: 2.7

5) Distance: $d=\sqrt{(4-3)^2+(7-1)^2}$

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: $y-5=-4(x-(-1))$

Step-by-step explanation:

The equation of a line is in the form

$y=mx+q$

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

$y=-2x+7$

So its slope is $m=-2$ . Therefore, the only line parallel to this one is the line which have the same slope, which is:

$y=-2x-3$

Since it also has $m=-2$

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: $d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32$

- Second diagonal: $d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32$

The diagonals are congruent, so this is a rectangle.

Given points A (0,1) and B (-2,5), the slope of the line is:

$m=\frac{5-1}{-2-0}=-2$

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

$m'=\frac{1}{2}$

And using the slope-intercept for,

$y-y_0 = m(x-x_0)$

Using the point $(x_0,y_0)=(7,1)$ we find:

$y-1=\frac{1}{2}(x-7)$

And re-arranging,

$y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5$

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

$x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7$

The distance between two points $A(x_A,y_A)$ and $B(x_B,y_B)$ is given by

$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

$d=\sqrt{(4-3)^2+(7-1)^2}$

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

$\frac{3}{3+5}=\frac{3}{8}$

of the distance between A and B.

To find the perimeter, we have to calculate the length of each side:

$d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6$

$d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2$

$d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2$

$d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6$

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

The area of a triangle is

$A=\frac{1}{2}(base)(height)$

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

$Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7$

$Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8$

So the area is

$A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8$

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

$y=\frac{1}{4}x-6$