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3 years ago

10

Suppose the point (3, 2) is changed to (3.1) on this rectangle. What other point must change so the figure remains a rectangle?

What is the area of the new rectangle

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

i need all the points

Step-by-step explanation:

since i do not have all the points for the rectangle....

3,2 goes to 3.1 the highest point to the most left will need to go down by so you will subtract the length between the 2 points to get your new area

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What’s a matrix and determinant?

Margaret [11]

Answer:

yan na po

Step-by-step explanation:

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7 0

3 years ago

Determine whether LINE AB and LINE CD are parallel, perpendicular, or neither. A(−1, −4) , B(2, 11) , C(1, 1) , D(4, 10) ( write

Luba_88 [7]

Step-by-step explanation:

Hey there!

The points of line AB are; (-1,-4) and (2,11).

Note:

Use double point formula and simplify it to get two eqaution.
Use condition of parallel lines, perpendicular lines to know whether the lines are parallel or perpendicular or nothing.

~ Use double point formula.

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

~ Keep all values.

$(y + 4) = \frac{11 + 4}{2 + 1} (x + 1)$

~ Simplify it.

$y + 4 = \frac{15}{3} (x + 1)$

$y + 4 = 5x + 5$

$5x - y + 1 = 0$

Therefore this is the equation of line AB.

Now, Finding the equation of line CD.

Given;

The points of line CD are; (1,1) and (4,10).

~ Using formula.

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

~ Keep all values.

$(y - 1) = \frac{10 - 1}{4 - 1} (x - 1)$

~ Simplify it.

$y - 1 = 3 x - 3$

$3x - y - 2 = 0$

Therefore, 3x - y- 2 = 0 is the eqaution of line CD.

Use condition of parallel lines.

m1= m2

Slope of equation (i)

$m1 = \frac{ - coeff. \: of \: x}{coeff \: of \: y}$

$m1 = \frac{ - 5}{ - 1}$

Therefore, m1 = 5

Slope of second equation.

$m2 = \frac{ - coeff \: .of \: x}{coeff \: .of \: y}$

$m2 = \frac{ - 3}{ - 1}$

Therefore, m2 = 3.

Now, m1≠m2.

So, the lies are not parallel.

Check for perpendicular.

m1*m2= -1

3*5≠-1.

Therefore, they aren't perpendicular too.

So, they are neither.

<em><u>Hope </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

6 0

3 years ago

7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and

Juliette [100K]

Answer:

A) JK and AB =

$[tex] \boxed{( \theta) = \f{12}{( \theta)} }$ So, =》 $\dfrac{10}{5} = \dfrac{1}{ \( \theta) }$ =》 $\( \theta) = \dfrac{5}{2}$ [/tex]

B) AABC

$[tex] \boxed{( \theta) = {1}{( \theta)} }$ So, $23}{-18=\\Ans } \boxed$

C) AABM:

$[tex] e \fbox{: }$ The value of k for which $\sin(jm) = \cos(x)$ is $\dfrac{\p}{2}$ ans[/tex]

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3 0

4 years ago

A line passes through the points (3,14) and (7,18) Write a linear function rule in terms of x and y for this line, Help! I might

MrRissso [65]

Answer:

y=1x+11

Step-by-step explanation:

1. Find the slope of the given points

$\frac{y2-y1}{x2-x1}$

$\frac{18-14}{7-3} =\frac{4}{4} =1$

2. Pick any point to find the y-intercept

(3, 14)

x=3, y=14, m=1

3. Substitute

y=mx+b

14=1(3)+b

14=3+b

14-3=b

11=b

4. Write the linear function rule

y=mx+b

y=1x+11

3 0

2 years ago

Read 2 more answers

find the equation of the line graph below in point slope form. show your work. (will make brainiest.)

frosja888 [35]

Answer:

y-2=4/3(x-2)

Step-by-step explanation:

Pick one coordinate and then plug the x value and the by value into the formula and then to find the slope and count up 4 and to the right 3.

7 0

4 years ago