All categories

2 years ago

This is due in 15 min i really need help

Mathematics

1 answer:

murzikaleks [220]2 years ago

5 0

Answer:

Parallel = lines with the same slope

Perpendicular = lines whose slopes are negative reciprocals (i.e. product is –1)

Step-by-step explanation:

The first part: circle b, c, d, f (1 ³/₄ is another way of saying ⁷/₄), and h (a horizontal line <em>y</em> = <em>b</em> has a slope of 0)

The second part: circle a, d (1 ⁴/₇ is another way of writing ¹¹/₇, the negative reciprocal of –⁷/₁₁), and e

Hope this helps! Have a great day!

You might be interested in

How will adding the value 90 affect the mean and median of the data set 1, 5, 5, 6, 9, 10?

Soloha48 [4]

I believe the answer is B for this question

6 0

3 years ago

Read 2 more answers

How do you make a equation with 2/3 X3

Marrrta [24]

Do you mean (2/3) times 3X or is x a multiplication symbol

8 0

3 years ago

Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

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

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

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

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

2 years ago

Someone please help!! I’ll give you Brainliest!

Valentin [98]

Answer:

x= 1

y= 2

(1,2)

x=2

y=3

(2,3)

x=3

y=4

(3,4)

5 0

3 years ago

Find an equation of the line l that passes through the point (x, y) = (−9, 2) and satisfies the condition. the x-intercept of l

spayn [35]

-9(2)+4=-16 you do -9 times 2 which is -18 then you subtract that to 4 which you get -16. hope this helps!

3 0

2 years ago