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Dmitriy789 [7]
3 years ago
11

Which of the following is a horizontal stretch of the parent function?

Mathematics
2 answers:
dlinn [17]3 years ago
8 0

Answer:

The correct option is 3.

Step-by-step explanation:

The parent function is

f(x)=|x|

The horizontal stretch and compression is defined as

g(x)=|bx|

If 0<b<1, then it is horizontal stretch and if b>1, then it is horizontal compression.

Graph 1 passing thought the point (1,-1).

-1=|b(1)|

This statement is false for any value of b, because the value of modulus can not be negative.

Graph 2 passing thought the point (1,2).

2=|b(1)|

2=b

Since b>1, therefore this graph represent the horizontal compression.

Graph 3 passing thought the point (2,1).

1=|b(2)|

\frac{1}{2}=b

Since 0<b<1, therefore this graph represent the horizontal stretch.

Thus, the correct option is 3.

melisa1 [442]3 years ago
6 0

Answer:

C.

Step-by-step explanation:

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How many solutions does a triangle with side lengths a = 4, A = 112º and b =<br>9 have?​
Ainat [17]

Answer:

This case has NO solutions.

Step-by-step explanation:

Notice that you are in a case of an obtuse triangle (one of its angles is larger than 90 degrees), the side opposite to the obtuse triangle is shorter than the side adjacent to the angle, so no actual triangle can be formed.

This can be found by simply trying to apply the Law of Sines to solve for the value of angle "B" opposite to side "b":

\frac{sin(A)}{a} =\frac{sin(B)}{b}\\sin(B)=\frac{b\,sin(A)}{a}\\sin(B)=\frac{9\,sin(112^o)}{4}\\\\sin(B)=2.086

As shown above, we get an impossible mathematical condition (also call an absurd), since the sine of an angle cannot give a value larger than 1 (one).

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7 0
3 years ago
Rationalize the denominator !!!
satela [25.4K]

Answer:

\frac{1\cdot \left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}\\

1\cdot \left(4+\sqrt{3}\right)=4+\sqrt{3}

\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right) = 13

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What is the answer please help!
Zigmanuir [339]

Answer:

the third one, c!

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The number of sod grasses needed is about one and half for the clients lawn

1 1/2 sod grasses

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Area of client's Lawn = Length * Width

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Learn more about rectangles here

brainly.com/question/25292087

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