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3 years ago

CAN YOU HELP ME FAST PLZ

Mathematics

2 answers:

Sedaia [141]3 years ago

6 0

Answer:

Options A, B, D and E are true.

Step-by-step explanation:

In the figure attached, trapezoid ABCD is dilated to create trapezoid A'B'C'D'.

Option A.

Length of AD = Distance of A from the origin + Distance of D from the origin

= 4 + 4

= 8 units

Option A is true.

Option B.

Length of A'D' = Distance of A' from the origin + distance of D' from the origin

= 2 + 2

= 4 units

Option B is true.

Option C.

As we have measured in the options A and option B

$\frac{AD}{A'D'}=\frac{8}{4}$

AD = 2× A'D'

So image is smaller than the pre image after dilation.

Option C is not true.

Option D.

Since coordinates of C and D are (2, 4) and (4, 0)

so the slope of CD = $\frac{y-y'}{x-x'}$

= $\frac{4-0}{2-4}$

= $\frac{4}{(-2)}$

= (-2)

Now we know coordinates of C' and D' are (1, 2) and (2, 0)

Then slope of C'D' = $\frac{2-0}{1-2}$

= $\frac{2}{-1}=(-2)$

Therefore, slopes of CD and C'D' are same.

Option D is True.

Option E.

Scale factor = $\frac{A'D'}{AD}=\frac{4}{8}$

= $\frac{1}{2}$

Option E is true.

Therefore, Options A, B, D and E are true.

Debora [2.8K]3 years ago

3 0

All of them are correct except the third option. The image is smaller than the pre-image after the dilations.

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1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

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$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

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