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2 years ago

How do I unsubscribe from here

Mathematics

2 answers:

Naddik [55]2 years ago

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Answer:

usub from where

Step-by-step explanation:

kap26 [50]2 years ago

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wdym by unsubscribe

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Kate has gone up to the chalkboard to do math problems 5 more times than Andre. Kate has gone up 11 times. Which equation repres

balandron [24]

B, because it says she has gone up 5 times Andre (a).

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2 years ago

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A graph titled Passing Statistics has pass attempts on the x-axis, and yards on the y-axis. A line goes through points (24, 127)

pishuonlain [190]

Answer:#2

Step-by-step explanation:

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2 years ago

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The hexagon GIKMPR is regular. The dashed line segments form 30 degree angles.

Gwar [14]

Answer:

The correct option is (D).

Step-by-step explanation:

It is given that GIKMPR is regular hexagon. It means it has 6 vertices.

Since the central angle is 360 degree. Therefore the central angle between two consecutive vertices is

$\frac{360^{\circ} }{6}=60^{\circ}$

It is given that the dashed line segments form 30 degree angles.

We have rotated the hexagon about O to map PQ to RF. Since P and R are consecutive vertices, therefore the angle between them is 60 degree.

The vertex R is immediate next to the vertex P in clockwise direction.

So if we rotate the hexagon at 60 degree clockwise about O, then we can maps PQ to RF.

$360^{\circ}-60^{\circ}=300^{\circ}$

Therefore we can also rotate the hexagon at 300 degree counterclockwise about O, then we can maps PQ to RF.

Therefore option D is correct.

8 0

2 years ago

In , QR = 4, RS = 15, and . In , VT = 8, TU = 32, and . Are the polygons similar? If they are, write a similarity statement and

iris [78.8K]

No because the scale factors are different. You can see that QR compared to VT the scale factor is 2, but in RS and TU it's different, so they are not similar.

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3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

2 years ago