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3 years ago

Help help help please

Mathematics

1 answer:

Anarel [89]3 years ago

5 0

Answer:

to add to the table, continue the arithmetic sequence in each column
to use the table, extend the third column to 45, or multiply the first row by 9

Step-by-step explanation:

The numbers in the first column are an arithmetic sequence with a common difference of 2. The next few numbers will be 6, 8, 10, ...

The numbers in the second column are an arithmetic sequence with a common difference of 3. The next few numbers will be 9, 12, 15, ...

The numbers in the third column are an arithmetic sequence with a common difference of 5. The next few numbers will be 15, 20, 25, ...

To use the table, Mark can extend each sequence until he has a row with 45 in the third column, or he can multiply or add rows to give a result of 45 in the third column. For example, multiplying the first row by 9 gives 18:27:45 meaning that Mark needs 18 liters of blue and 27 liters of yellow to make 45 liters of green paint.

You might be interested in

The sides of a triangle have lengths of x, x + 4, and 20. If the longest side is 20, which of the following values of x would ma

GuDViN [60]

Answer:

12<x<16

Step-by-step explanation:

if c^2 > a^2 + b^2

then the triangle is obtuse

since 20 is the longest side

20^2 > x^2 + (x+4)^2

simplify

400 > x^2 + (x+4)(x+4)

FOIL

400 > (x^2 ) + (x^2 + 4x+4x+16)

combine like terms

400 > (2x^2 + 8x+16)

divide by 2

400/2 > 2x^2 /2 + 8x/2 + 16/2

200 > x^2 + 4x +8

subtract 200 from each side

0> x^2 + 4x +8-200

0> x^2 +4x-192

Factor

0 > (x-12) ( x+16)

using the zero product property

0> x-12 0 > x+16

12>x -16>x

x must be greater than 12

we know the longest side is 20

x+4 < 20

subtract 4

x< 16

x>12 and x < 16

12<x<16

5 0

3 years ago

For which of the following compound inequalities is there no solution?

meriva

Answer:

Step-by-step explanation:

$3m - 12 > 30 \text{ and } -6m\geq 24$

$\boxed{3m - 12 > 30 \wedge -6m\geq 24}$

$m>14 \wedge m\leq -4$

There's no solution. It is the first one already. There is no number that is both greater than 14 and less than or equal to -4. That is no solution because there's no $m$ that satisfy the compound inequality.