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Katen [24]
3 years ago
11

When does a parabola open upward

Mathematics
2 answers:
RUDIKE [14]3 years ago
8 0

Answer:

If the x is squared, the parabola is vertical (opens up or down). If the y is squared, it is horizontal (opens left or right). If a is positive, the parabola opens up or to the right. If it is negative, it opens down or to the left.

Step-by-step explanation:

Sidana [21]3 years ago
4 0

Answer:

y=ax^2+bx+c

When "a" is positive, opens upward.

When "a" is negative, opens downward.

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What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE
Nutka1998 [239]

Answer:

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),  \left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

Step-by-step explanation:

Let points D, E and F have coordinates (x_D,y_D),\ (x_E,y_E) and (x_F,y_F).

1. Midpoint M of segment DF has coordinates

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).

2. Midpoint N of segment EF has coordinates

\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0
3 years ago
21x-5=7+17x <br> Write it down step by step please
Nat2105 [25]

Answer: x=3

Step-by-step explanation: Step 1: Simplify both sides of the equation.

21x−5=7+17x

21x+−5=7+17x

21x−5=17x+7

Step 2: Subtract 17x from both sides.

21x−5−17x=17x+7−17x

4x−5=7

Step 3: Add 5 to both sides.

4x−5+5=7+5

4x=12

Step 4: Divide both sides by 4.

\frac{4x}{4} = \frac{12}{4}

So Final Answer: <u>x=3</u>

<u />

<u>If I Helped, Please Mark Me As Brainliest, Have A Great Day :D</u>

4 0
3 years ago
Read 2 more answers
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
3 years ago
Read 2 more answers
What is one more than -5
Cerrena [4.2K]
All real numbers above -4
3 0
3 years ago
Read 2 more answers
What is the value of v?<br>(picture attached)<br>PLEASE ANSWER ASAP!!!​
azamat

Answer:

v=12

Step-by-step explanation:

180-12v+13v-50+38=180

168+v=180

v=12

6 0
3 years ago
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