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AlladinOne [14]
3 years ago
11

Parallel lines r and s are cut by two transversals, parallel lines t and u.

Mathematics
2 answers:
mote1985 [20]3 years ago
8 0

Answer:

The correct answer would be Angle 5 and Angle 13.

Step-by-step explanation:

I was taking the test when I commented on this question later, I ended up getting the question right.

Darya [45]3 years ago
5 0

Answer:

ghhhhhh

Step-by-step explanation:

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Write the equation of the line in slope-intercept form.
Ymorist [56]

Answer:

y = 1/40x + 20

Step-by-step explanation:

Slope-intercept equation format:

y = mx + b

m = slope/ gradient/ constant of variation/ rate of change

b = y-intercept

To find the slope, use the formula:

\frac{y_2-y_1}{x_2-x_1}

Find  two random points:

1 - (0.5, 40) , 2 - (1, 60)

Plug in:

\frac{60-40}{1-0.5}

Solve:

\frac{20}{0.5}

Divide:

\frac{1}{40}

The y-intercept is when the x-coordinate is equal to 0.

The y-intercept is always the beginning.

So observing the graph, the first point is (0, 20)

The x-coordinate is 0.

Y intercept = 20.

Plug in both slope and y-intercept:

y = 1/40x + 20

8 0
2 years ago
Write an equation in point slope form for the line that passes through (-5,6) and (-3,-9).
Finger [1]

\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\qquad
(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-9})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-6}{-3-(-5)}\implies \cfrac{-9-6}{-3+5}\implies -\cfrac{15}{2}


\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-6=-\cfrac{15}{2}[x-(-5)]\implies y-6=-\cfrac{15}{2}(x+5)
\\\\\\
y-6=-\cfrac{15}{2}x-\cfrac{75}{2}\implies y=-\cfrac{15}{2}x-\cfrac{75}{2}-6\implies y=-\cfrac{15}{2}x-\cfrac{87}{2}

8 0
3 years ago
At 5pm the total rainfall is 3cm. At 11 pm the total rainfall is 13cm.What is the mean hourly rainfall?
krok68 [10]

Answer:

2.16 cm

Step-by-step explanation:

5 through 11 = 6 hours

total of cms  = 13

13/6 = 2.16

7 0
3 years ago
Use the Quadratic formula to find all real zeros of the 2nd degree polynomial
schepotkina [342]

Answer:

  x ∈ {-5, -1}

Step-by-step explanation:

Here's the solution using the quadratic formula:

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-24\pm\sqrt{24^2-4\cdot 4\cdot 20}}{2\cdot 4}\\\\=\dfrac{-24\pm\sqrt{576-320}}{8}=\dfrac{-24\pm\sqrt{256}}{8}\\\\=\dfrac{-24\pm 16}{8}=-3\pm 2=\{-5,-1\}

The real zeros are -5 and -1.

_____

There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.

4 0
3 years ago
Please i need help asap!! ​
chubhunter [2.5K]

Answer:

(-4,-1) is the answer to this.

7 0
2 years ago
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