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Svetradugi [14.3K]
3 years ago
12

Number 17 and 16 plz help

Mathematics
1 answer:
fgiga [73]3 years ago
8 0
Hi. Okay, #16 says write a decimal between 0.5 and 0.75. Then write it as a fraction in simplest form and as a percent.

write a decimal = I will pick 0.15 - now we need to write it as a fraction since the 5 is in the hundredths place our fractions would be:

15/100 = 3/20 (simplest form)

Now to change .15 to a percent we simply move the decimal two places to the right and add a % sign so. 15 = 15%

Question 17
How would you write 43 3/4% as a decimal

For this one, it is a lot easier than you think. The answer is 43.75. Let me tell you how...

To change 3/4 to a decimal, you are merely dividing the 4 into 3.00, hence the .75 and then just bring the 43 over. With problem like this that end in 1/4, 1/2, 3/4 just think of 4 quarters---1 quarter =.24, 2 quarters = .50, 3 quarters =.75

Question 16 answer is: .15,  15/100=3/20 and 15%
Question 17 answer is: 43.75

I hope this helps. If you have any questions, please don't hesitate to ask.

Take care,
Diana


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Question 1. The sticker price of a picture frame is $10 and the sales tax is 10%. What is the total price of the picture frame?
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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)
strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

y=a(x-h)^{2} +k

where

(h,k) is the vertex

if a>0 -------> the parabola open upward (vertex is a minimum)

if a -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> f(x)=(x-2)^{2}

This is a vertical parabola open upward

the vertex is the point (2,0)

therefore

The function f(x)=(x-2)^{2}  does not have a vertex on the y-axis

<u>case B)</u> f(x)=x(x+2)

f(x)=x(x+2)=x^{2}+2x

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+1=x^{2}+2x+1

Rewrite as perfect squares

f(x)+1=(x+1)^{2}

f(x)=(x+1)^{2}-1

the vertex is the point (-1,-1)

therefore

The function f(x)=x(x+2) does not have a vertex on the y-axis

<u>case C)</u> f(x)=(x-2)(x+2)

f(x)=(x-2)(x+2)=x^{2}-2^{2}

f(x)=x^{2}-4

the vertex is the point (0,-4)

The x-coordinate of the vertex is equal to zero

therefore

The function f(x)=(x-2)(x+2) has a vertex on the y-axis

<u>case D)</u> f(x)=(x+1)(x-2)

f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+2= x^{2} -x

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+2+0.25= x^{2} -x+0.25

f(x)+2.25= x^{2} -x+0.25

Rewrite as perfect squares

f(x)+2.25= (x-0.50)^{2}

f(x)=(x-0.50)^{2}-2.25

the vertex is the point (0.5,-2.25)

therefore

The function f(x)=(x+1)(x-2) does not have a vertex on the y-axis

<u>the answer is</u>

f(x)=(x-2)(x+2)

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