Which undefined terms are needed to define parallel lines?

Mathematics

1 answer:

Shalnov [3]3 years ago

6 0

Coplanar and line are the undefined terms needed to define parallel lines.

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PLEASE NEED HELP NOW!!!!!!!❤️

alexira [117]

Answer:

A. 2564 cm^2

Step-by-step explanation:

There are two triangles (Which both are the same) and three different rectangles.

1. Let's solve the area of the triangles first!

Formula: bh(1/2)

10(26)(1/2)

130

2. Now, let's multiply by 2 since there's two of them:

130(2) = 260

3. So, we got 260. Let's find the area of the three rectangles!

Formula: bh

Rectangle A: 28(36) = 1008

Rectangle B: 10(36) = 360

Rectangle C: 26(36) = 936

4. Add the area of the triangles and the rectangles!

260 + 1008 + 360 + 936 = 2564

So the surface area is 2564 cm^2

Hope this helps! <3

4 0

3 years ago

PLEASE PLEASE HELP !!!!!!!!

nataly862011 [7]

Answer:c

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Code. -3x + 4y = -8 8x - y = 16 Solving systems by substitution

LiRa [457]

Answer:

The solution to the system of equations is $\displaystyle \big(\frac{56}{29}, -\frac{16}{29}\big)$ .

Step-by-step explanation:

We are given a system of equations:

$\displaystyle \left \{ {{-3x+4y=-8} \atop {8x-y=16}} \right.$

We need to solve these by substitution, so we need to solve one equation for a variable and then substitute the value of that variable into the other equation.
After doing this and solving for the opposite variable, we need to insert this into the original equation and solve for the initial variable.
Therefore, to solve one of the equations, we will solve it to put it in slope-intercept form and solve for y. The easier equation to work with is equation two.

$\displaystyle 8x - y = 16\\\\-y = -8x + 16\\\\\frac{-y}{-1}=\frac{-8x+16}{-1}\\\\y = 8x - 16$

Now, we've solved for y. So, we can substitute this into either equation and solve for x.

$\displaystyle -3x + 4(8x -16)=-8\\\\-3x + 32x - 64 = -8\\\\29x - 64 = -8\\\\29x = 56\\\\\frac{29x}{29}=\frac{56}{29}\\\\x = \frac{56}{29}$

Now, we substitute our value for x into one of the original equations and solve for y.

$\displaystyle 8\big(\frac{56}{29}\big)-y=16\\\\-y = 16 - 8\big(\frac{56}{29}\big)\\\\-y=\frac{16}{29}\\\\\frac{-y}{-1}=\frac{\frac{16}{29}}{-1}\\\\y = -\frac{16}{29}$