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3 years ago

The graph of a quadratic function contains the points

Mathematics

1 answer:

masya89 [10]3 years ago

3 0

Answer:

Shawn is correct.

Step-by-step explanation:

Let the quadratic function is g(x) = a(x - h)² + k

Here (h, k) is the vertex of the parabola.

Since this parabola passes through (0, 0), (1, 9) and (-1, 9), axis of symmetry is x = 0 and the vertex is (0, 0).

Therefore, equation of the parabola will be,

g(x) = a(x - 0)²+ 0

g(x) = ax²

for a point (1, 9) which lies on the graph,

9 = a(1)²

a = 9

g(x) = 9x² (here a > 1)

Therefore, f(x) is vertically stretched by a factor of 9 to form g(x).

Shawn is correct.

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Tania bought 4 more pounds of pears than Wilma. Together, Tania and Wilma bought 18 pounds of pears.

horrorfan [7]

The equation would be (x+4)+x=18. X stands for the amount of pears Wilma bought, and (x+4) is the amount of pears Tania bought (because she bought four more pounds than Wilma). (X+4)+x=18 2x+4=18 (now you subtract four from both sides) 2x=14 (to get x by itself, divide both sides by 2) X=7 Wilma bought 7 pounds of pears :3

7 0

3 years ago

Read 2 more answers

Find the area of sector DEF (show work please)

Whitepunk [10]

Answer:

240

Step-by-step explanation:

360-120=240

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3 years ago

Answer will get the brainliest

tia_tia [17]

I agree with him that’s the answer

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3 years ago

What is the area of a triangle with vertices (3,0) (9,0) (7,6)

Nezavi [6.7K]

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (3,0) (9,0) (7,6). So

$A = \pm 0.5 \left|\begin{array}{ccc}3&0&1\\9&0&1\\7&7&1\end{array}\right|$

$A = \pm 0.5(3*0*1+0*1*7+1*9*7-0*7*1-0*9*1-3*1*7)$

$A = \pm 0.5*(63-21)$

$A = \pm 0.5*42 = 21$

The area of the triangle is of 21 units of area.

5 0

3 years ago

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0

3 years ago