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2 years ago

Help me with this plz

Mathematics

2 answers:

Feliz [49]2 years ago

8 0

Answer is Y=3 Times

Lisa [10]2 years ago

5 0

The answer is the first one

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SOMEONE PLEASEEEEE HELPPPPP MEEEEE

Hatshy [7]

Answer:

(5,3)

There is no explanation for this solution.

8 0

3 years ago

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Please answer fast The vertices of a rectangle are located at the coordinates (1, 4), (1, 5), (6, 5),

hammer [34]

Answer:

See below ~

Step-by-step explanation:

Drawing the rectangle (Refer attachment)

Horizontal sides

There are two x-values present : 1 and 6
Find the difference
6 - 1 = 5
The horizontal sides of the rectangle are 5 units long

Vertical sides

Two y-values are present : 4 and 5
Find the difference
5 - 4 = 1
The vertical sides of the rectangle are 1 unit long

Perimeter

2(Horizontal side + Vertical side)
2(5 + 1)
2(6)
12
The perimeter of the rectangle is 12 units

3 0

1 year ago

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love history [14]

Answer:

36 in²

Step-by-step explanation:

Area of the figure

Area (small rectangle) left + Area (big rectangle) right
3 x (6 - 4) + 5 x 6
3 x 2 + 30
6 + 30
36 in²

3 0

2 years ago

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-7(a - 3) = 11 - 7a what’s the answer

natali 33 [55]

Answer:

the statement is false there is no solution

Step-by-step explanation:

step 1 -7(a-3)=11-7a

step 2 -7a+21=11-7a

step 3 cancel out the -7a's

step 4 you are left with 21=11 which does not work so it is false

8 0

3 years ago

Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

Step 1

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

Step 2

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

Step 3

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

Step 4

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

Step 5

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

Step 6

Segment EF, segment FG, segment GH, and segment EH are congruent.

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

Step 7

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

4 0

3 years ago

Read 2 more answers