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ICE Princess25 [194]

4 years ago

6

Which transformation does not preserve size ? Dilation rotation translation line reflection

2 answers:

Rudiy274 years ago

8 0

Dilation. Dilation is the only transformation to change size

anzhelika [568]4 years ago

7 0

Dilation. Dilation augments or shrinks a shape, while rotation and line reflection keep the shape the same size.

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A pastry recipe calls for 3 cups of flour to make 8 servings. How many cups of flour are needed to make 6 servings?

natima [27]

The answer is A. If you divide 3/8 which equals .375. Then you multiply .375 by 6 to get 2.25!

:D Hope this helps!

5 0

4 years ago

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Which of the following represents a geometric sequence?

ziro4ka [17]

The answer is IV because I took the test

3 0

3 years ago

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This figure is made up of a quadrilateral and a semicircle.

vodomira [7]

Here the figure is made up of a quadrilateral and a semi circle.

ABCD is the quadrilateral here. We will find the sides of the quadrilateral by using the distance formula.

If (x₁, y₁) and (x₂, y₂) are two points given, then the distance between two points by using distance formula is,

$d =\sqrt({ x_{1}-x_{2})^2 +( y_{1} -y_{2} )^2}$

The co-ordinate of A is (-1,2) and co-ordinate of B is (-2,-1).

So the length of side AB = $\sqrt{(-1-(-2))^2+(2-(-1))^2}$

= $\sqrt{(-1+2)^2+(2+1)^2}$ (As negative times negative is positive)

= $\sqrt{(1)^2+(3)^2} =\sqrt{1+9} =\sqrt{10}$

The co-ordinate of C is (4,-3) and D is (5,0)

The length of side CD

= $\sqrt(4-5)^2+(-3-0)^2}$

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

= $\sqrt{1+9} =\sqrt{10}$

So the sides AB and CD are equal.

The length of side AD

= $\sqrt{(-1-5)^2+(2-0)^2}$

= $\sqrt{(-6)^2+(2)^2} =\sqrt{36+4} =\sqrt{40}$

The length of side BC

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

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

= $\sqrt{(-6)^2+(2)^2}=\sqrt{36+4} =\sqrt{40}$

So the lengths of the sides AD and BC are equal.

So the quadrilateral is a rectangle whose length is $\sqrt{40}$ and width is $\sqrt{10}$ .

Area of a rectangle = length × width

= $(\sqrt{40}) (\sqrt{10})$

= $\sqrt{(40)(10)}=\sqrt{400}$

= $20 unit^2$

Now the diameter of the semicircle is the side AD = $\sqrt{40}$

So, the radius of the semi-circle = $\frac{\sqrt{40}}{2}$

= $\frac{\sqrt{(4)(10)}}{2}$

= $\frac{(\sqrt{4})(\sqrt{10})}{2}$

= $\frac{2\sqrt{10}}{2}$ = $\sqrt{10}$

Area of semi-circle = $\frac{1}{2} \pi r^2$ , where r is the radius.

= $\frac{1}{2} \pi (\sqrt{10})^2$

= $\frac{1}{2} \pi (10)$

= $\frac{(\pi)(10)}{2}$

= $\frac{10\pi}{2} = 5\pi$ = $15.7 unit^2$ ( Approximately taken to the nearest tenth)

Total area of the figure = $(20+15.7) unit^2$ = $35.7 unit^2$

We have got the required answer.

Option a is correct here.

3 0

3 years ago

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Convert 21.3 feet per minute to yards per minute.

OverLord2011 [107]

Answer:

7.1 yard per min

Step-by-step explanation:

7 0

3 years ago

Consider the equation:

Vladimir [108]

Answer:

3+2 = 22 +30= 59 1/2

(x + c)2 = d or (x – c)2 = d. = (c * d)(x) = 4x^3 + 18x^2 - 10x

X=1+2

Step-by-step explanation:

sorry if wrong was every confused

6 0

3 years ago