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Sunny_sXe [5.5K]
3 years ago
10

How do you represent real-world problems involving ratios and rates with tables and graphs?

Mathematics
1 answer:
yaroslaw [1]3 years ago
6 0

Answer:

we can study the relationship between the two given variables or quantities

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Which list shows the absolute values in order from least to greatest?
Maurinko [17]
Your answer is going to be A, B, and D :)
7 0
4 years ago
Read 2 more answers
Help me I really need to get this done
posledela

Answer:

It’s the second one

Step-by-step explanation:

6 0
3 years ago
2v2 -5v +3 = 0 how did I do this?
anzhelika [568]
2v2 -5v +3 = 0 \\ \\\Delta = b^{2}-4ac = (-5)^{2}-4*2*3= 25-24=1\\ \\v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{5- \sqrt{1 }}{2*2}=\frac{5- 1}{4}= \frac{4}{4}=1\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{5+ \sqrt{1 }}{2*2}=\frac{5+1}{4}= \frac{6}{4}=\frac{3}{2}=1\frac{1}{2}


8 0
4 years ago
Which 2 values of x are roots of the polynomial below?
Alexxx [7]

Given:

The given polynomial is:

f(x)= 3x^{2}-3x+1

To find the roots of the given polynomial.

To find the roots we have to take f(x) = 0

So,

3x^{2} -3x+1 = 0

Formula

By quadratic formula, the root of the equation ax^{2} +bx+c = 0 is,

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

Now,

Putting, a=3, b=-3, c=1 we get,

x = \frac{3+\sqrt{3^{2}-4(3)(1) } }{(2)(3)} and \frac{3-\sqrt{3^{2}-4(3)(1) } }{(2)(3)}

x = \frac{3+\sqrt{-3} }{6} and x=\frac{3-\sqrt{-3} }{6}

Hence,

The values of the roots of the given polynomial arex=\frac{3+\sqrt{-3} }{6} and x=\frac{3-\sqrt{-3} }{6}

Hence, Option A and F are the correct answer.

5 0
3 years ago
The absolute value function, shifted 1 unit downward and 5 units to the right.
neonofarm [45]

The absolute value function is:

f(x)=\lvert x\rvert

To translate it we need to remember the following rules:

Using this table we conclude that the function we are looking for is:

f(x)=\lvert x-5\rvert-1

5 0
1 year ago
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