Which lists the fractions

Can anyone here help me out?

Flauer [41]

Answer:

A)15x+30=weekly income

B)8 hours

Step-by-step explanation:

15x+30= weekly income

The 15 multiplies because it says per hour

The 30 is constant because she only earns it once (per week)

15x+30=150

minus 30 from both sides

15x=120

divide by 15

x=8

6 0

3 years ago

If the property tax on a 300,000 Is 2,220 what is the tax on a 100,000 home

ikadub [295]

Answer:

SO first you have to divise 300,000 by 2,220 and then multiply it by 100,000.To get 13636363.6364.SO the tax on a 100k home would be about 13636363.6364

Step-by-step explanation:

6 0

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:

1)

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 $m=-2$

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.

3)

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,

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

4)

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:

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

5)

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

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

6)

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.

7)

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

8)

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

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

9)

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

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

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

$m'=-4$

Using the slope-intercept form,

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

And using the point

$(x_0,y_0)=(-1,5)$

we find:

$y-5=-4(x-(-1))$

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0

3 years ago

How many solutions does the system have?Explain

valkas [14]

There are NO SOLUTIONS.

Substitute the first equation for y in the second equation. You can see it does not work. 6 does not equal 8. Therefore there are no solutions.

6 0

3 years ago

The angle which is less than 90 degrees is called?

BlackZzzverrR [31]

Answer:

An acute angle is an angle less than 90°

5 0

3 years ago

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