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aksik [14]
3 years ago
10

Find and round to the nearest cent. 35% of $246.55

Mathematics
1 answer:
katrin [286]3 years ago
5 0
Hello,
35%=0.35
To find the answer:
$246.55×0.35=$86.2925 or $86.3. As a result, 35% of $246.55 is $86.4. Hope it help!
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PLEASE HELP WILL BE GIVING BRAINLIEST!!!<br><br> -13+13r≥-2(11-3r)-7
wlad13 [49]

Answer:

r\geq\frac{-16}{7}

Step-by-step explanation:

-13+13r\geq-22+6r-7

bring all r and number one oppsite sides

13r - 6r \geq -22-7+13

7r \geq -16

r \geq \frac{-16}{7}

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sumaya is reading a book with 288 pages she has already read 90 pages she plans to read 20 more pages each day until she finshes
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Answer:


Step-by-step explanation:

you could create an equation which would be 288= 90 + 20(x)

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3 years ago
Which fractions are equivalent to the following?
LiRa [457]
It would be B. if that helps

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3 years ago
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423 times 28 in grid method
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Answer:

Step-by-step explanation:

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3 years ago
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)

6 0
3 years ago
Read 2 more answers
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