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Anuta_ua [19.1K]
2 years ago
8

WILL MARK BRAINILIEST

Mathematics
1 answer:
olganol [36]2 years ago
7 0

Answer:

these are consecutive interior angles

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What is the slope if you have (2,1) and (4,4)?<br><br><br>*Ineed this quickly as possible!-Thx
Schach [20]

(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{4}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{2}}}\implies \cfrac{3}{2}

7 0
2 years ago
Read 2 more answers
: Matteo has only $62.50 to spend at a soccer store. He wants to buy a jersey that costs $27.50 including tax, and socks that co
Anastasy [175]

Answer:

Step-by-step explanation:

$62.50                              $35.00/ $3.50 = 10

-$27.50                               x=10

-------------

$35.00

Matteo can buy a maximum of 10 socks.

3 0
3 years ago
HELPPPPPP MEEEEE!!!!!!
Kazeer [188]

Answer:

It's a

Step-by-step explanation:

The speed goes up, and so does the time

then the speed stays the same but the time keeps going

then the time keeps going and the speed goes down

3 0
3 years ago
Looking at the two quadratic functions below (1 &amp; 2), answer the following questions.
algol [13]
Part A:

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).



Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).



Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).



Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).
7 0
3 years ago
Helppp plzzzzzz!!!!!!
melisa1 [442]

Answer: a

Step-by-step explanation:

7 0
3 years ago
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