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2 years ago

What is the value of this expression?

Mathematics

1 answer:

solmaris [256]2 years ago

4 0

$\frac{11}{12}-\frac{1}{3} +\frac{2}{6} = \frac{11}{12}-\frac{1}{3} +\frac{1}{3} =\frac{11}{12}$

Answer: 11/12

Ok done. Thank to me :>

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13. Based on the pattern, what are the next two terms of the sequence?

Anastasy [175]

Answer:

C. 22, 27

Step-by-step explanation:

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3 years ago

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PLZZZ HELP I WILL MARK BRAINIEST

charle [14.2K]

Answer: 47.5 ft^2

Step-by-step explanation: First find the area of the rectangle using the formula length*height. This is 10*4 or 40.

Then find the area of the bottom triangle using the formula (base*height)/2. This will be (3*3)/2, which simplifies to 9/2, which is 4.5.

Then find the area of the top triangle using the same formula, which will be (2*3)/2, which simplifies to 6/2, which is 3.

Add all of these together. 40+3+4.5=47.5.

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3 years ago

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3 years ago

What is 7.07106781187 rounded to the nearest hundredth?

Keith_Richards [23]

7.07 is the right answer to 7.07106781187 rounded to the nearest hundredth

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4 years ago

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

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3 years ago