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Hoochie [10]
3 years ago
15

Parallelogram PQRS is shown on the coordinate grid. Parallelogram PQRS is dilated with the origin as the center of dilation usin

g the rule (x, y) → (0.75x, 0.75y) to create parallelogram P'Q'R'S'.
Which statement is true?

A) Parallelogram P'Q'R'S' is larger than parallelogram PQRS, because the scale factor is greater than 1.

B) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is less than 1.

C) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is greater than 1.

D) Parallelogram P'Q'R'S' is larger than parallelogram PQRS, because the scale factor is less than 1.
Mathematics
1 answer:
Gwar [14]3 years ago
7 0

Answer:

B) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is less than 1.  

Step-by-step explanation:

Given that Parallelogram PQRS is shown on the coordinate grid. Parallelogram PQRS is dilated with the origin as the center of dilation using the rule (x, y) → (0.75x, 0.75y) to create parallelogram P'Q'R'S'.

This means that sides of PQRS would be 0.75 times the sides of the original PQRS

Or PQRS will be larger

B) Parallelogram P'Q'R'S' is smaller than parallelogram PQRS, because the scale factor is less than 1.  

is the right answer

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Answer: A: 3/5

Step-by-step explanation:

He made 9 shots

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The ratio will be 9/15, which can be simplified to 3/5

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5 0
3 years ago
Find the exact value of each trigonometric function for the given angle θ.
Kay [80]

Answer:

\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since \theta=240, it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}

\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}

\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}

\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}

\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2

\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}

Therefore, \sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.

8 0
3 years ago
How do I solve this?
meriva
Hello,

If 3x-1>=0 then

x>=1/3
|3x-1\=3x-1
2*|3x-1|=18
2*(3x-1)=18
6x-2=18
6x=20
x=10/3

else
x<1/3
|3x-1|=-(3x-1)
2*|3x-1|=18
2(-(3x-1))=18
-6x+2=18
-6x=16
x=-8/3
endif


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mrs_skeptik [129]

Answer:

if number was represented by x it would be x+5

Step-by-step explanation:

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Step-by-step explanation:

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