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3 years ago

2/5 + k/4 = 9/10 (solve with work)

Mathematics

2 answers:

Sliva [168]3 years ago

8 0

Remark

The easiest way to solve this is to change the fractions into decimals.

2/5 = 0.4

9/10 = 0.9

k/4 = 1/4 k = 0.25 k

0.4 + 0.25k = 0.9 Subtract 0.4 from both sides.

0.25k = 0.9 - 0.4

0.25k = 0.5 Divide by 0.25

k = 0.5 / 0.25 Use your calculator for this if you don't know how to do it.

k = 2 <<<< answer.

Alla [95]3 years ago

6 0

The LCD of 2/5 + k/4 = 9/10 is 20. Applying this LCD, we get:

8 + 5k = 18. Subtracting 8 from both sides: 5k = 10. Then k = 10/5, or k = 2.

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3 years ago

Read 2 more answers

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

$(x-h)^2+(y-k)^2=r^2$

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

8 0

2 years ago

Please help me

dusya [7]

3(4x - 2) = 2(6x - 3)

12x - 6 = 12x - 6

Add 6 on both sides

12x = 12x

Divide by 12 on both sides

x = x

That means this has infinite solutions. Your answer is B.

4 0

2 years ago

Read 2 more answers