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2 years ago

What are the zeroes of the function y=3(x-a)(x+b) in terms of a and b?

Mathematics

1 answer:

juin [17]2 years ago

4 0

Answer:

a, -b

Step-by-step explanation:

When x is a, the function becomes

$f(a)=3(a-a)(a+b)$

as you see, a-a is 0, and as your whole function is multiplied by this, it makes it 0.

$f(-b)=3(-b-a)(-b+b)$

As you see -b+b is 0, and once again, this would make the whole function 0.

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2 years ago

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MrRissso [65]

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2 years ago

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2 years ago

The length of a picture is 15.75 inches shorter than twice the width. If the perimeter of the picture is 106.5 inches, find its

Vladimir79 [104]

Answer:

Length = 45.37 in

Width = 30.56 in

Step-by-step explanation:

Let the width of the picture be 'w' inches.

Given:

Perimeter of the picture is, $P=106.5\ in$

Length is 15.75 inches shorter than twice the width. So, length is:

$l=2w-15.75$

Now, we know that the perimeter of a rectangular shape is given as the sum of all of its side. So, the picture is also of rectangular shape and hence the perimeter is given as:

$P=2l+2w$

Plug in the values of 'l' and 'P' and solve for 'w'. This gives,

$106.5=2w-15.75+2w\\106.5=4w-15.75\\106.5+15.75=4w\\4w=122.25\\w=\frac{122.25}{4}=30.56\ in$

Therefore, the width of the rectangular picture is 30.56 inches.

Length of the picture is, $l=2(30.56)-15.75=61.12-15.75=45.37\ in$

So, length of the picture is 45.37 inches and width is 30.56 inches.

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2 years ago