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Charra [1.4K]
3 years ago
7

Help me …… please ! I Just need help

Mathematics
1 answer:
frozen [14]3 years ago
7 0

Answer:

(2,-2)

Step-by-step explanation:

y=-4x+6

y=3x-8

plug the y's in for each other to get:

-4x+6=3x-8   <----- solve from there

+8            +8

-4x+14=3x

+4x         +4x

14=7x

/7     /7

x=2

now plug the x in to get y (in any problem)

y=3(2)-8

y=6-8

y=-2

<em>(2,-2)</em>

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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
Javier walks up to a tank of water that can hold up to 10 gallons. When it is active, a drain empties water from the tank at a c
kap26 [50]

Hello, Bellaroyalball, your answer would be,

4 1/2 minutes before Jada arrived

I hope this helps you - Astralyradele

5 0
2 years ago
How can you rewrite the expression (8-5i)^2 in the form a+bi?
otez555 [7]
(a-b)^2 = a^2-2ab+b^2

(8-5i)^2 = 8^2-2(8)(5i)+(5i)^2

= 64-80i+25i^2

i^2=-1

So

= 64-80i+25(-1)

=64-25-80i

= <em><u>39 - 80i</u></em>

which is your answer :)
8 0
4 years ago
Read 2 more answers
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
What is 7/6 ÷ 4 in simplest form
serg [7]
\frac{7}{6} \div 4= \frac{7}{6} \times \frac{1}{4} = \frac{7\times 1}{6\times 4} = \frac{7}{24}

7/24 is your answer :)
7 0
3 years ago
Read 2 more answers
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