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3 years ago

7

Larry is designing the floor of a building. The design of the floor will be in the shape of a rectangular grid consisting of n^3

columns by n^2 rows of tiles. Which expression shows the number of tiles he plans to use for the floor?

1 answer:

kondaur [170]3 years ago

7 0

I don't know this answer sorry boy/girl

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Solve each equation by completing the square. If necessary, round to the nearest hundredth.

Ksenya-84 [330]

$1.)\\ \\ b^2+10b=75\\ \\ b^2+10b-75=0\\ \\\Delta = b^{2}-4ac = 10^{2}-4*1*(75)=100+300=400 \\ \\\sqrt{\Delta }=\sqrt{400}=20$

$b_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-10-20}{2}=\frac{-30}{2}=-15\\ \\b_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-10+20}{2}=\frac{10}{2}=5$

$2.)\\ \\ n^2-20n=-75\\ \\ n^2-20n+75=0\\ \\\Delta = b^{2}-4ac = (-20)^{2}-4*1*75=400-300=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$n_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{20-10}{2}=\frac{10}{2}=5\\ \\n_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{20+10}{2}=\frac{30}{2}=15$

$3.)\\ \\t^2+8t-9=0\\ \\\Delta = b^{2}-4ac = 8^{2}-4*1* (-9)=64+36=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$t_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-8-10}{2}=\frac{-18}{2}=-9\\ \\t_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{-8+10}{2}=\frac{2}{2}=1$

$4.)\\ \\m^2-2m-8=0\\ \\\Delta = b^{2}-4ac = (-2)^{2}-4*1* (-8)=4+32=36\\ \\\sqrt{\Delta }=\sqrt{36}=6$

$m_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{2-6}{2}=\frac{-4}{2}=-2\\ \\m_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{2+6}{2}=\frac{8}{2}=4$

$5.)\\ \\ v^2+4v-2=0\\ \\\Delta = b^{2}-4ac = 4^{2}-4*1* (-2)=16+8=24\\ \\\sqrt{\Delta }=\sqrt{24}= \sqrt{4*6}=2\sqrt{6}$

$v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-4-2\sqrt{6}}{2}=\frac{2(-2-\sqrt{6}}{2}= -2-\sqrt{6}\approx -2-2,45\approx -4,45\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{-4+2\sqrt{6}}{2}=\frac{2(\sqrt{6}-2}{2}= \sqrt{6}-2 \approx 2,45-2\approx 0,45$

3 0

3 years ago

Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10.

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

That will be a square with a diagonal of 20

side length of 20sin45

and area of (20sin45)² = 200 units²

prove it you say?

Area of a rectangle is base times height

A = bh

With a radius of 10, the diagonals of any rectangle inscribed will be 20 units

20² = b² + h²

h = $\sqrt{400 - b^2}$

A = bh

A = b $\sqrt{400 - b^2}$

Area will be maximized when the derivative is set to zero

dA/db = $\sqrt{400 - b^2}$ - b²/ $\sqrt{400 - b^2}$

0 = $\sqrt{400 - b^2}$ - b²/ $\sqrt{400 - b^2}$

b²/ $\sqrt{400 - b^2}$ = $\sqrt{400 - b^2}$

b² = 400 - b²

2b² = 400

b² = 200

b = $\sqrt{200}$

h = $\sqrt{400 - b^2}$

h = $\sqrt{400 - \sqrt{200}^2 }$

h = $\sqrt{200}$

A = bh

A = $\sqrt{200}$ • $\sqrt{200}$

A = 200 units²

5 0

3 years ago