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erma4kov [3.2K]
3 years ago
6

If two sides of a triangle measure 19 cm and 34 cm, check all possible values for the third side.

Mathematics
1 answer:
Blizzard [7]3 years ago
6 0

Answer:

Step-by-step explanation:

The length of the third side of a triangle must always be between (but not equal to) the sum and the difference of the other two sides.

19 + 34 = 53

34 - 19 = 15

from 16 cm  to 52 cm

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A 1.50 liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25oC. Without change in t
cricket20 [7]
The answer is 4.5 l

To calculate this, we will use the Boyle's law using constant k, pressure P, and volume V:

 PV = k

We can forget temperature since it is constant.

We have:

P1V1 = k

P2V2 = k

So, P1V1 = P2V2

P1 = 3 atm

V1 = 1.5 l

P2 = 1 atm

V2 = ?

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P1V1 = P2V2

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4.5 atm/l = 1 atm * V2

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3 years ago
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The dimensions of a rectangle are:
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Answer:-24x -10

Step-by-step explanation:

Dimensions:-6x-5 and -6x

Perimeter:2(l+b)

=2(-6x-5 + -6x)

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Hope it helps ❤️

7 0
3 years ago
a machine takes 2.8 hours to make 9 parts .at that rate how many parts can the machine make in 28.0?
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Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and
NNADVOKAT [17]

Answer:

The vector equation of the line is \overrightarrow {r}=(11j-8k)+t(4i-4j+6k) and parametric equations for the line are x=4t, y=11-4t, z=-8+6t.

Step-by-step explanation:

It is given that the line passes through the point (0,11,-8) and parallel to the line

x=-1+4t

y=6-4t

z=3+6t

The parametric equation are defined as

x=x_1+at,y=y_1+bt,z=z_1+ct

Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.

From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.

The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are

x=4t

y=11-4t

z=-8+6t

Vector equation of a line is

\overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}

where, \overrightarrow {r_0} is a position vector and \overrightarrow {v} is cosine of parallel vector.

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Therefore the vector equation of the line is \overrightarrow {r}=(11j-8k)+t(4i-4j+6k) and parametric equations for the line are x=4t, y=11-4t, z=-8+6t.

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No table that shows that is posted with the question.
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