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2 years ago

Find the values for k so that the intersection of x = 2k and 3x + 2y = 12 lies in the first quadrant. PLEASSSSEEEE

Mathematics

2 answers:

Dovator [93]2 years ago

8 0

Answer:

0< k < 2

Step-by-step explanation:

x = 2k

3x + 2y = 12

Substitute the first equation into the second equation

3(2k) +2y = 12

6k+2y = 12

Subtract 6k from each side

2y = 12 -6k

Divide by 2

y = 6 - 3k

For this to be in the first quadrant y > 0

6 - 3k > 0

6 > 3k

Divide by 3

2 > k

The lower limit is when it reaches the axis or when k=0

0 <k <2

Checking for x

x = 2k

If 0<k <2 then it will be in the first quadrant

gulaghasi [49]2 years ago

7 0

Answer:

k = any value between 0 and 2

Step-by-step explanation:

We can start by considering the graph of 3x + 2y = 12. If you convert this line into point - slope form it will have a negative slope, and hence will extend infinitely in the negative direction. Therefore if we calculate the x - intercept, k will be bound by 0 to the x - intercept if their intersection lies in the first quadrant.

3x + 2y = 12,

3x + 2(0) = 12,

3x = 12,

x = 4 - this is the x - intercept

As x = 4 is the x - intercept, k can be any value between 0 and 4, and will lie in the first quadrant. But remember that we have the equation " x = 2k. " k will only be between 0 and 4 is the equation is " x = k. "

- Therefore k will instead be any positive value between 0 and 2.

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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

astraxan [27]

ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
${i}^{2} = - 1$
to simplify the above expression.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}$

This implies that,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1$

8 0

2 years ago

Read 2 more answers

Use properties to rewrite the given equation. Which equations have the same solution as x + + x = – x plus StartFraction 2 Over

Contact [7]

Answer:

The answer is A B and D

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B. 18x + 20 + 30x = 15 – 6x

C. 18x + 20 + x = 15 – 6x

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E. 12x + 30x = –5