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Basile [38]
3 years ago
5

If y varies directly as x, what is the value of y in these ordered pairs? (8, 4) and (4, y)

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
6 0
So if we see the first one (8,4) x=8 and y=4

And we see the 2nd one (4,y)  x=4 and y=?

We notice that the value of x its been halved. So y varies with x so y value also has to be halved so y= 2
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The equation below shows the relationship between the number of apples purchased, a, and the total cost
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Answer:

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a (5) = c (4)

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20 (total) divided by the 5 apples is $4.

6 0
3 years ago
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WILL GIVE BRAINLIEST
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Answer:

\sqrt{11}

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2 years ago
This question is from trignometry of class 9.​
Ne4ueva [31]

Answer:

  4.2π cm ≈ 13.19 cm

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3 years ago
What is the difference in the range of temperature between the two years
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3 years ago
Simplify the rational expression. State any restrictions on the variable n^4-11n^2+30/ n^4-7n^2+10
djyliett [7]
To factor both numerator and denominator in this rational expression we are going to substitute n^{2} with x; so n^{2} =x and n ^{4} =  x^{2}. This way we can rewrite the expression as follows:
\frac{n^{4}-11n^{2} +30 }{n^{2}-7n^{2} +10 } =  \frac{ x^{2} -11x+30}{ x^{2} -7x+10}
Now we have two much easier to factor expressions of the form a x^{2} +bx+c. For the numerator we need to find two numbers whose product is c (30) and its sum b (-11); those numbers are -5 and -6. (-5)(-6)=30 and -5-6=-11.
Similarly, for the denominator those numbers are -2 and -5. (-2)(-5)=10 and -2-5=-7. Now we can factor both numerator and denominator:
\frac{ x^{2} -11x+30}{ x^{2} -7x+10} = \frac{(x-6)(x-5)}{(x-2)(x-5)}
Notice that we have (x-5) in both numerator and denominator, so we can cancel those out:
\frac{x-6}{x-2}
But remember than x= n^{2}, so lets replace that to get back to our original variable:
\frac{n^{2}-6 }{n^{2}-2 }
Last but not least, the denominator of rational expression can't be zero, so the only restriction in the variable is n^{2} -2 \neq 0
n^{2}  \neq 2
n \neq +or- \sqrt{2}
5 0
3 years ago
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