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2 years ago

Geometry.... ://...help please!

Mathematics

1 answer:

bekas [8.4K]2 years ago

8 0

Answer: it would be (9,13)

Step-by-step explanation:

if you go up you have to add 3 unites to Y which is (6) and add 5 unites to X which is (8) (hope this helped) :)

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A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

2 years ago

The quotient of w and 5 is at least 25

const2013 [10]

I think the answer is 5 because 25 divided by 5=5

8 0

3 years ago

Read 2 more answers

What is The equation of a vertical line passing through the point (-5,-1)

mihalych1998 [28]

The equation of a vertical line passing through the point (-5,-1) is x = -5

3 0

2 years ago

Read 2 more answers

Please help!!!!!!<br> I need help

murzikaleks [220]

Answer:

Step-by-step explanation: What you want to do is find the grid point of each for example D is -4,-4 than you want to add those like B is 8,-6 so you would do -4+-8 and -4+-6 so you get 2,-10 for the distance of DB

6 0

2 years ago

Lydia graphed ADEF at the coordinates D (-2,-1), E (-2, 2), and F (0,0). She thinks ADEF is a right triangle. Is Lydia's asserti

Setler79 [48]

The true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

<h3>Right triangles </h3>

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

D = (-2,-1)
E = (-2,2)
F = (0,0)

<h3>Slopes</h3>

Start by calculating the slopes of lines DF and EF using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{DF} = \frac{0 + 1}{0 +2}$

$m_{DF} = \frac{1}{2}$

Also, we have:

$m_{EF} = \frac{0 -2 }{0+2}$

$m_{EF} = \frac{-2 }{2}$

$m_{EF} = -1$

For the triangle to be a right triangle, then the calculated slopes must be opposite reciprocals.

i.e.

$m_1 = -\frac{1}{m_2}$

By comparison, the slopes of both lines are not opposite reciprocals.

Hence, the true statement is (c)

Geometry.... ://...​help please!

Geometry.... ://...help please!