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allochka39001 [22]
3 years ago
9

What are the x-intercepts of the parabola?

Mathematics
2 answers:
ELEN [110]3 years ago
4 0

Answer:

(4,0) and (5,0)

Step-by-step explanation:

The solutions of a parabola is when the parabola crosses the x- intercept.

Natasha_Volkova [10]3 years ago
3 0

Answer:

(5, 0) and (4, 0)

Step-by-step explanation:

According to the graph shown, the parabola passes trough the points (4,0) and (5,0). Therefore, they are the x-intercepts of the parabola.

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Could someone help me rnnn?
GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola:  y=3x^2-4

Step-by-step explanation:

Given:

  • y-intercept of parabola: -4
  • parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:  y=a(x-h)^2+k

(where (h, k) is the vertex and a is some constant)

Substitute point (0, -4) into the equation:

\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}

Substitute point (-2, 8) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}

Substitute point (1, -1) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}

Equate to find h:

\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}

Substitute found value of h into one of the equations to find a:

\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}

Substitute found values of h and a to find k:

\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}

Therefore, the equation of the parabola in vertex form is:

\implies y=3(x-0)^2-4=3x^2-4

So the vertex of the parabola is (0, -4)

5 0
2 years ago
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Encrypt the word PLACE using an alphabetic Caesar shift cipher with shift 13 ( mapping A and N)​
Vladimir [108]
PLACE

P > C

L > Y

A > N

C > P

E > R

CYNPR
6 0
3 years ago
if two angles are complementary then the sum of their measure is 90. if the sum of the measures of two angles is 90 then both of
Tju [1.3M]
Yes, both angles are acute because all complementary angles are acute.
6 0
3 years ago
100 POINTS TO RIGHT AWNSER!!!!
makvit [3.9K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000586

——————————

The answer is option D)  r < 5  or  r > – 1.

I'm going to graph each inequality below on a number line.


A)  r > 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B)  r > 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r

Numbers between – 1  and  5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C)  r < 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D)  r < 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

Now all real numbers are included in the union. So this is the right choice.


Answer:  option D)  r < 5  or  r > – 1.


I hope this helps. =)

6 0
3 years ago
Solve the system of equations displayed in the picture.
bagirrra123 [75]

The answer is D) (-1,-12)

Reason:

1) substitute 2x-10 for y in y=4x-8

2) 2x-10=4x-8

3) -2x-10=-8

4) -2x=2

5) x=-1

6) substitute -1 for x in y=2x-10

7) y=(2)(-1)-10

8) y=-12

Answer

x=-1 and y=-12

7 0
3 years ago
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