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3 years ago

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Which is not an equation of the line going through (3, –6) and (1, 2)?

2 answers:

uysha [10]3 years ago

7 0

Answer:

^correct

Step-by-step explanation:

vodka [1.7K]3 years ago

6 0

I do belive that it is C.

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How to tell if its identifying functions is <br> a function or not

Svetach [21]

Answer:

Step-by-step explanation:

Determining whether a relation is a function on a graph is relatively easy by using the vertical line test. If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function.

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3 years ago

Can someone plz help me it’s due at 11:30 plz help!!!

arsen [322]

Well it is b I’m 80% sure

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3 years ago

Through: (6, 2), perpendicular to<br> y = -x-5 <br> in slope intercept form

True [87]

Answer:

y = x-4

Step-by-step explanation:

y = -x-5

Slope of line is -1.

Slope of perpendicular to line is 1.

Point-slope equation for line of slope 1 that passes through (6,2):

y-2 = 1(x-6)

Rearrange equation to slope-intercept form:

y = 1(x-6) + 2

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6 0

3 years ago

What’s 10 - 3.587 equal

Pavel [41]

Instead of brainly you could just open your calculator.

10-3.585 = 6.415

7 0

3 years ago

Read 2 more answers

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

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

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

3 years ago