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3 years ago

Which lines are perpendicular to the line y – 1 = (x+2)? Check all that apply.

Mathematics

2 answers:

goblinko [34]3 years ago

6 0

Answer:

none of them

Step-by-step explanation:

The given line is, written in a more convenient way:

$y = x + 3$

I moved the (-1) to the right hand side, it became (+1) and summed with (+2) gave the (+3).

As you can see, the slope of your line is +1.

Now what do we know from the theory? We know that two lines are perpendicular if one's slope is the negative periodical of the other's.

In your case this is very easy to calculate. The negative reciprocal of +1 is -1.

The slopes of the five cases given as options are as follows:

1) -3

2) +3

3) -3

4) +1

5) -3

None of those lines is perpendicular to the original given line.

Ilya [14]3 years ago

5 0

Answer:

none of them

Step-by-step explanation:

Two lines are perpendicular when satisfy the next equation: m1*m2 = -1, where m1 and m2 are the slopes o the lines.

line 1:

y – 1 = (x+2)

y = x + 3

slope of line 1 = 1

line 2:

y + 2 = –3(x – 4)

y + 2 = -3*x + 12

y = -3*x + 10

slope of line 2 = -3

m1*m2 = 1*(-3 ) = -3

They are not perpendicular

line 3:

y − 5 = 3(x + 11)

y − 5 = 3*x + 33

y = 3*x + 38

slope of line 3 = 3

m1*m3 = 1*3 = 3

They are not perpendicular

line 4:

y = -3x –

slope of line 4 = -3

m1*m4 = 1*(-3 ) = -3

They are not perpendicular

line 5:

y = x – 2

slope of line 5 = 1

m1*m5 = 1*1 = 1

They are not perpendicular

line 6:

3x + y = 7

y = -3x + 7

slope of line 6 = -3

m1*m6 = 1*(-3 ) = -3

They are not perpendicular

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3 years ago

Laney builds a tower with wooden cubes. The bottom cube's edges are 8 centimeters long. The middle cube's edges are 2 centimeter

Mazyrski [523]

Answer: The total volume of the the cubes in the tower is 792 cubic centimetres (792 cm³)

Step-by-step explanation: We shall call the volume of the cube at the bottom VB, the volume of the cube at the middle VM, and the volume of the cube at the top VT. The tower is made up of cubes at different levels and at the bottom the cube measures 8 centimetres. The cube at the middle measures 2 cm less than the bottom cube, hence middle cube equals 8 minus 2 which equals 6 cm. The top cube measures 2 cm less than the middle cube, hence the top cube equals 6 minus 2 which equals 4 cm. The volume of each cube is given as;

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The length of a cube measures the same on all sides, that is, length, width and height. The length on all sides therefore of the bottom cube is 8 cm. The volume equals;

VB = 8³

VB = 512 cm³

The length on all sides of the middle cube is 6 cm (measures 2 cm shorter than the bottom cube). The volume of the middle cube equals;

VM = L³

VM = 6³

VM = 216 cm³

The length on all sides of the top cube is 4 cm (measures 2 cm shorter than the middle cube). The volume of the top cube equals;

VT = L³

VT = 4³

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From the calculations shown, the total volume of the cubes in the tower is given as;

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Total volume is 792 cubic centimetres.

8 0

3 years ago

Read 2 more answers

PLEASE HELP !!!! I'm stuck

irina [24]

The answer is c not D you are welcome

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3 years ago

Divide. Write the quotient in the lowest terms 12 divided by 1 and 1 over 5

IgorLugansk [536]

you turn 1 and 1 over 5 into 6 over 5. then divide.

12 divided by 6 over 5 equals 10

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3 years ago

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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