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garik1379 [7]
2 years ago
10

Make a table of values for the following equation:

Mathematics
2 answers:
mars1129 [50]2 years ago
7 0

Answer:

Step-by-step explanation:

We choose arbitrary values for x and in each case calculate y = 2x + 5.

If x = 0, y = 5; if x = 3, y = 11, and so on.  

How?  when x = 3, y = 2(3) + 5 = 6 + 5 = 11.

The table, with rows and columns suitable to this problem, would look like:

x  y

0  5

3  11

10 25

and so on

djverab [1.8K]2 years ago
7 0

Answer:

Step-by-step explanation:

Plug in anything for x and you will know what y is. For example if x is 0, y is 5. x is 12, y is 29.

A table would be like:

x - 0, 12, 10

y - 5, 29, 25

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6 0
3 years ago
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Hi please i nedd help with these questions .
Vlad1618 [11]

Answer:

Step-by-step explanation:

#1:

\frac{18m^2u}{16n^3v^2} \div \frac{24m}{15nu^3}*\frac{8n^2v^3}{30m^3u}

division sign means that we flip the fraction

\frac{18m^2u}{16n^3v^2} * \frac{15nu^3}{24m}*\frac{8n^2v^3}{30m^3u}

now we can multiply all the constants together and all variables

m: (m^2)/(m*m^4) = (m^2)/(m^4) = 1/(m^2)\\u: (u * u^3)/(u) = (u^4)/(u) = u^3\\n: (n*n^2)/(n^3) = 1\\v: (v^3) / (v^2) = v\\(18*15*8)/(16*24*30) = \frac{3}{16}

now we can combine all the parts

\frac{3u^3v}{16m^2}

#2:

\frac{24a^2b^3c}{9bc^3}\div\frac{4a^5bc^3}{27a^3b^2c}

\frac{24a^2b^3c}{9bc^3}*\frac{27a^3b^2c}{4a^5bc^3}

a: (a^2 * a^3)/(a^5) = 1\\b: (b^3 *b^2)/(b*b) = b^3\\c: (c*c)/(c^3*c^3)= 1/(c^4)\\(24*27)/(9*4)= 18

\frac{18b^3}{c^4}

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Answer:

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Step-by-step explanation:

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