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vladimir1956 [14]
3 years ago
11

I am a fraction equivalent to 1/2. The

Mathematics
2 answers:
denis-greek [22]3 years ago
4 0

Answer:

5/10

Step-by-step explanation:

5/10 = 1/2

5 + 10 = 15

tankabanditka [31]3 years ago
4 0

Answer:

5/10 is the fraction.

Step-by-step explanation:

5/10 = 0.5

1/2 = 0.5

5 + 10 = 15

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Marcus said the product of 7/8 × 1 3/4 is 9/4.<br><br> How can you tell that his answer is wrong?
Paraphin [41]
Gravity simply does not allow that
8 0
2 years ago
Find the probability of drawing a green card, not replacing it, and then drawing another green card.
IRISSAK [1]

Answer:

1/15

Step-by-step explanation:

This is probably too late but I'll do it anyway.

There are 10 cards in all.

of the 10, 3 are green.

So your chance of drawing the first green is 3/10

Now you have 9 cards in total left.

Two of them are green

Your chance of drawing a green card again is 2/9

Your total chance of drawing 2 green cards is

2/9 * 3/10 = 6/90

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7 0
2 years ago
An expression involves subtracting two numbers from a positive number. Under what circumstances with the value expression be neg
Svetach [21]
So (positive integer)-x-y can be negative
so to get a negative number as your result/answer, x+y must be greater than (positve integer) so  for example


lets say (positive integer)=8
x+y must be greater than 8 to make the equation negative so
example
x=5
y=4
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8-9=-1
7 0
3 years ago
Alicia has five times as many dimes as she has quarters. Combined the dimes and quarters total to $9.75. How many dimes does Ali
olasank [31]
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3 0
2 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
2 years ago
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