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mamaluj [8]
3 years ago
15

A pair of parallel lines is cut by a transversal, as shown below: A pair of parallel lines is cut by a transversal. The exterior

angle made on the right by the intersection of the upper parallel line and the transversal is labeled as p. The exterior angle made on the left by the intersection of the lower parallel line and the transversal is labeled as q. Which of the following best represents the relationship between angles p and q? p = 180 degrees − q q = 180 degrees − p p = 2q p = q

Mathematics
2 answers:
Furkat [3]3 years ago
6 0

Answer:

b: Lines p and q are parallel because alternate exterior angles are congruent

Step-by-step explanation:

STatiana [176]3 years ago
3 0

Because they are alternate exterior angles, they are the same.

So the best relationship from the given is:  p = q

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Determine if the statement is always, sometimes or never true. Equilateral triangles are isosceles triangles.
tia_tia [17]
Never .. isoscels is unequal. I'm pretty sure
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3 years ago
Are these equations infinite solution, no solution, or one solution?
Karo-lina-s [1.5K]

Answer:

1) one solution

2) Infinitely many solution

3) no solution

Step-by-step explanation:

The first equation is -14-8x=-2(-3x+7)

Let us expand the right hand side to get:

-14-8x=-2*-3x+-2*7

We multiply -14-8x=6x-14

We group similar terms to get:

-8x-6x=-14+14

Combine similar terms to get:

-14x=0

Divide through by -14 to get:

x=0

2) The second equation is 3+5x=5(x+2)-7

We expand to obtain:

3+5x=5x+5*2-7

Multiply to get:

3+5x=5x+10-7

This gives us

3+5x=5x+3

5x=5x

1=1

There is infinitely many solutions

3) The third equation is: 36-7x=-7(x-5)

We expand to get:

36-7x=-7x+35

Group similar terms;

-7x+7x=35-36

0=-1

This statement is not true so there is no solution.

5 0
3 years ago
Find the cost of painting a cubical block of side length 3.2 cm, if painting costs $5 per cm2.
SashulF [63]
<h3><u>Answer</u>:- </h3>

\longrightarrow \sf \$ \: 307.2

<h3><u>Solution:-</u></h3>

Since painting is done on outer surface of cubical block we will have to find its surface area which is given by-

\green{ \underline { \boxed{ \sf{Surface \: area  \:of  \:Cube = 6a^2}}}}

  • <u>where a is side of the cube </u>

\begin{gathered}\\\quad \sf Surface \: area  \:of  \:Cube = 6\times(3.2)^2  \\\end{gathered}

\begin{gathered}\\\implies \quad \sf  6\times 3.2\times3.2   \\\end{gathered}

\begin{gathered}\\\implies \quad \sf  61.44 cm^2  \\\end{gathered}

Now,

\maltese \sf Cost  \:of  \:painting \: 1 \: cm² = \$ 5

\sf Cost \: of \: painting \: 61.44  \:cm²=61.44×5

\sf\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\$ 307.2

\leadstoThus it would cost $ 307.2 to paint the entire cubical block.

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I dont really know how to do the work but i guess 1/125 is the simplest from to do 0.008 into a fraction 
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(9/16)*(1/9) = 1/16 . . . . . . 1st selection
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