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2 years ago

When lines are parallel...

Mathematics

1 answer:

Otrada [13]2 years ago

3 0

Answer:

1) Congruent

2) Supplementary

3) Congruent

4) Congruent

Step-by-step explanation:

1) The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent .

2) Formally, we can say that if two lines are parallel, then consecutive interior angles are supplementary. We refer to this as the consecutive interior angles postulate.

3) When the lines are parallel, the corresponding angles are congruent . When two lines are cut by a transversal, the pairs of angles on one side of the transversal and inside the two lines are called the consecutive interior angles.

4) The Alternate Exterior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate exterior angles are congruent.

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PLZ ANSWER ASAP WILL GIVE 50PTS if u explain and mark brainliest!!!!!!!

Vinil7 [7]

Answer:

Step-by-step explanation:

$\frac{5q+5p}{2} = 3+4p\\5q+5p = 6+8p \\5q = 6 +3p \\q = \frac{6+3p}{5} ( a=3,b=6,c=5)$

2. q-p = 3p +5

q = 4p+5

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p = (q-5)/4

3. y = 2a -b $x^{2}$

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2 years ago

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The probability is 1/1024.

Each tetrahedron has a 1/4 chance of landing on 3, since there are 4 sides and only one of them is marked 3.

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The average speed of a car is three times as fast as the average speed of a cyclist. to travel 225 km, the cyclist requires 5 h

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3 years ago

Help! what is the right answer?

egoroff_w [7]

Answer:

Step-by-step explanation:

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2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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2 years ago