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2 years ago

Linear inequalities What is the solution to: -1/3 a + 4 ≤ 0

Mathematics

1 answer:

kati45 [8]2 years ago

8 0

Step-by-step explanation:

-1/3 a + 4 ≤ 0

3*(-1/3a) + 4*3 ≤ 0*3

-a + 12 ≤ 0

12 ≤ a

12 ⩾a

a⩾12

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Lee wants to cut this piece of canvas into two rectangles that are 3×2 and 3×5. He wants the sum of the areas of the two small r

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2 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

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Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$