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Lynna [10]
4 years ago
15

Write a real word problem for the equation 4xy=60 . Then solve

Mathematics
1 answer:
Readme [11.4K]4 years ago
5 0
The word problem would be four x multipied by a number equals 60

4(xy)=60
(4*x)*(4*y)=60
then u plg 0 in for x or y and solve for each variable
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24 thriller movies out of 36 DVDs and 10 thriller movies out of 15 DVDs
disa [49]

Answer:

Ok??

Step-by-step explanation:

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3 years ago
What’s GD<br> Answers<br> a. 20<br> b. 3<br> c. 15<br> d. 12
djverab [1.8K]

Answer:

B. 3

Step-by-step explanation:

16 : 4 = 12 : |GD|

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4 years ago
If α and β are the zeroes of the polynomial 6y 2 − 7y + 2, find a quadratic polynomial whose zeroes are 1 α and 1 β .
ollegr [7]

Answer:

2y^2-7y+6=0

Step-by-step explanation:

We are given that \alpha and \beta are the zeroes of the polynomial 6y^2-7y+2

y^2-\frac{7}{6}y+\frac{1}{3}

We have to find a quadratic polynomial whose zeroes are 1/\alpha and 1/\beta.

General quadratic equation

x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes

We get

\alpha+\beta=\frac{7}{6}

\alpha \beta=\frac{1}{3}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2

\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}

\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3

Substitute the values

y^2-(7/2)y+3=0

2y^2-7y+6=0

Hence, the quadratic polynomial whose zeroes are 1/\alpha and 1/\beta is given by

2y^2-7y+6=0

4 0
4 years ago
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Step-by-step explanation:

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6 0
4 years ago
Read 2 more answers
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Readme [11.4K]

Answer:

53

Step-by-step explanation:

So all you have to do is multiply 1.06 * 50 = 53

3 0
4 years ago
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