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dalvyx [7]
3 years ago
12

Sally sold a decorated flowers that cost the same amount each plus a dozen roses for $28.All together she sold $73 in flowers.Ho

w much was each decorated flower?
Mathematics
1 answer:
Dominik [7]3 years ago
5 0
To begin you must subtract the cost of the roses from the total cost of the flowers sold so $73-$28.  Then divide the remainder by the number of decorated flowers she sold to get the cost per each decorated flower.  I do not see in your question the number of decorated flowers she sold.  I see "a" so perhaps she sold one???
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5x + 3 = 7x – 1. Find x<br><br> a. 1/3<br> b. ½<br> c. 1<br> d. 2
weqwewe [10]

Answer:

Step-by-step explanation:

First you need to get like items on the same side(put the variables on one side and the non-variables on the other).  Remember anything you do on one side of the equal, you have to do on the other side, too.

5X+3=7X-1   I'm going to move the -1 first, but you don't have to as long as you keep the equation equal.

5X+3+1=7X-1+1    We know that +1-1 is zero and so is -1+1.  So

5X+3+1=7X      We don't need to keep that zero.

5X+4=7X         Now, we want to move the 5X over.

5X-5X+4=7X-5X    Simplify and get rid of the zero.

4=2X             Last, we have to divide to get the 2 away from the X.

4/2=2X/2            Not everyone know that slash mean divide, :-).

2=X

3 0
3 years ago
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
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