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Serggg [28]
3 years ago
10

Can u show me common core way how to get the answer to this problem 4 1/5. ×4 3/8

Mathematics
1 answer:
AVprozaik [17]3 years ago
5 0
I'm not quite sure what you mean by common core way but the way to get the answer the question is this:

1.) You must covert the whole numbers into fractions.

4 \frac{1}{5}  \: will \: become \:  \frac{21}{5}
(multiply the denominator by the whole number, which in this case is 4, then add the answer to the numerator).

4 \frac{3}{8 \:  }  \: will \: become \:  \frac{35}{8}
(same steps will apply as before except in this case there is a different denominator and numerator).

There is no need to make the denominators the same number since it is multiplication and not addition or subtraction (the same can be done with division expect there is one crucial difference which I can explain if you want, in the comments).

2.) Next set up the equation to multiply:

\frac{21}{5}  \times \frac{35}{8}
5 and simplify (also note that the only way you can't simplify is across)

\frac{21}{5}  \times  \frac{7}{8}
3.) Now multiply across:

NUMERATOR
21 × 7 = 147

DENOMINATOR
5 × 8 = 40

It will come together to be 147/40 or 3 27/40
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the axis of symmetry of a quadratic equation is x = –3. if one of the zeroes of the equation is 4, what is the other zero?
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ANSWER

The other zero is

x =  - 10

EXPLANATION

The axis of symmetry serves as the midpoint of the two zeroes.


We were given that the axis of symmetry of the quadratic equation is
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We were also given that, one of the zeroes of the quadratic equation is
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Let the other zero of the quadratic equation be
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\frac{x_1 + x_2}{2}  = axis \: of \: symmetry

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\frac{p + 4}{2}  =  - 3



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p + 4 =  - 6



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p =  - 6 - 4

p =  - 10




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3 years ago
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Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.
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Answer:

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Step-by-step explanation:

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\pm \frac{\sqrt{13}}{3}=\sec(\theta)

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

\sec(\theta)=-\frac{\sqrt{13}}{3}

\cos(\theta)=-\frac{3}{\sqrt{13}}.

I don't see this answer as I'm going to rationalize the denominator.

\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}.

\cos(\theta)=-\frac{3\sqrt{13}}{13}.

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Step-by-step explanation:

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