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3 years ago

If p is parallel to q, which pair of angles must be congruent?

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

5 0

Vertical angles are congruent whether the lines are parallel or not.
Corresponding angles are congruent.
Alternate interior and alternate exterior angles are congruent.
1 & 4, 2 & 3, 3 & 6, 4 & 5, 1 & 8, 2 & 7, 5 & 8, 6 & 7, 3 & 7, 1 & 5, 2 & 6, 4 & 8
I think I got all of them!

yuradex [85]3 years ago

4 0

When two lines are parallel and are intersected by a third line,

Then the pairs of corresponding angles are congruent.

The pairs of Alternate interior angles are congruent.

More names can be given to these angles.

But when we consider all corresponding and alternate angles, all angles are covered.

In the given figure

<2 = <6, <4 = <8, <1 = <5 and <3= <7 are all corresponding angles. ..(1)

Also < 4 = <5 & <3 = <6 are alternate interior angles. ...(2)

If we compare Statement (1) & statement (2)

We get

<1 = <5 = <8 = <4

<3 = <7 = <6 = <2

So all red dotted angles are equal

Also all black dotted angles are equal.

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MrMuchimi

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3 years ago

How do u graph x+y=-2

sdas [7]

Answer:

The answer is

$y = - x - 2$

Step-by-step explanation:

The problem is

$x + y = - 2$

First we have to bring x to the other side se we subtract x from both sides

$x + y = - 2 \\ - x \: \: \: = - 2 - x$

Then we have to put it in order so the answer is

$y = - x - 2$

4 0

3 years ago

I need these three and they shouldn’t be too hard.

kondor19780726 [428]

Answer:

i dont know

Step-by-step explanation:

sorry my man my school is still just going into these kind of questions ):

4 0

3 years ago

Solving a Two-Step Matrix Equation Solve the equation:

Cloud [144]

Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

$3x_{1} + 2x_{2} + 1 = 2$

-Second equation:

$5x_{1} + 5x_{2} + 2 = -3$

-Choose one of the two following equations, which I choose the first one, then you solve for $x_{1}$ by isolating

$3x_{1} + 2x_{2} + 1 = 2$

-Subtract $1$ to both sides:

$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

$3x_{1} + 2x_{2} = 1$

-Subtract $2x_{2}$ to both sides:

$3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1$

$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

-Combine like terms:

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

$\frac{5}{3}x_{2} + \frac{11}{3} = -3$

-Subtract $\frac{11}{3}$ to both sides:

$\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}$

$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

-After you have the value of $x_2$ , substitute for $x_{2}$ onto this equation, which is $x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$ :

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

-Multiply $-\frac{2}{3}$ and $-4$ :

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

$x_{1} = \frac{8 + 1}{3}$

-Since both $\frac{1}{3}$ and $\frac{8}{3}$ have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

$x_{1} = \frac{8 + 1}{3}$

$x_{1} = \frac{9}{3}$

$\boxed {x_{1} = 3}$

5 0

3 years ago

10. Using the following arithmetic sequence, find the 29th term: –36, –32, –28, –24.

Mashcka [7]

–36, –32, –28, –24 This is an arithmetic sequence because each term has the same difference from the preceding term, called the common difference, d...

-32--36=-28--32=-24--28=4 So 4 is d, the common difference.

The sequence of any arithmetic sequence has the form:

a(n)=a+d(n-1), a=first term, d=common difference, n=term number...in this case we have:

a(n)=-36+4(n-1)

a(n)=-36+4n-4

a(n)=4n-40 so the 29th term is:

a(29)=4(29)-40

a(29)=116-40

a(29)=76

...

distance=velocity * time

d=vt we want to find t so

t=d/v and in this case:

t=234/70

t=(210+24)/70

t=3hr+(60*24)/70

t=3hr+20min+34sec so

t≈3hr 20min

...

This is an arithmetic sequence...100,150,200...

The sum of an arithmetic sequence will always be the average of the first and last terms times the number of terms....

the rule for the sequence is:

a(n)=a+d(n-1), a(n)=100+50d-50, a(n)=50n+50

Now we know the nth term is 50n+50, and we also know the first term is 100 so:

s(n)=n(100+50n+50)/2 and we want to know the sum of the first 10 terms so

s(10)=10(100+500+50)/2

s(10)=$3250

...

The first two terms are 2 and 4 so:

a(n)=2+2(n-1)

a(n)=2+2n-2

a(n)=2n

a(10)=20

...

You could do synthetic or long division, but you also could just use the fact that the factor being (x+8) should indicate a zero for the function when x=-8. If f(x) could be divided by (x+8) the value of y(-8) would equal zero, however calculating y(-8)=-10 so that would be the remainder if you did the division.

8 0

3 years ago