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Find an equation in standard form of the parabola passing through the points (1, -2), (2,-2), (3,-4)

Yuliya22 [10]

The standard form of a parabola is y=ax²+bx+c
use the three given points to find the three unknown constants a, b, and c:
-2=a+b+c............1
-2=4a+2b+c......... 2
-4=9a+3b+c...........3
equation 2 minus equation 1: 3a+b=0..........4
equation 3 minus equation 2: 5a+b=-2.........5
equation 5 minus equation 4: 2a=-2, so a=-1
plug a=-1 in equation 4: -3+b=0, so b=3
Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
the parabola is y=-x²+3x-4

double check: when x=1, y=-1+3-4=-2
when x=2, y=-4+6-4=-2
when x=3, y=-9+9-4=-4
Yes.

8 0

2 years ago

A dog walks at a speed of 3mi/hr for 2 hours. How far (distance) did the dog walk? *

Nikolay [14]

Answer:

6 miles

Step-by-step explanation:

After one hour the dog will have walked three miles after two hours the dog will have walked three more miles for a total of 6 miles.

6 0

2 years ago

Helpppp meeee fassttt pleaseeee

Yuki888 [10]

Answer:

y = -1/3x + 3 2/3

Step-by-step explanation:

step 1. find the slope of the given line

y2-y1/x2-x1 2 - 4/ 3 + 3 = - 1 / 3

take the opposite of that and there is your slope

step 2. plug in the values to the following equation using the given point:

y - y1 = m ( x - x1 ) y - 4 = - 1/3 ( x + 1 )

step 3. solve for y

y-4 = -1/3(x+1)

-1/3(x) and -1/3(1) ----> y-4 = -1/3x-1/3

y-4+4 = -1/3x-1/3+4 ----> y = -1/3x + 3 2/3

<em>hope this helps. if you need more help lmk in the comments. i am in algebra two so you can trust me. Happy holidays</em>

3 0

2 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago

1.

Yanka [14]

Answer:

6 units down

Step-by-step explanation:

Transformation is the movement of a point from its original position to a new position. If an object is transformed, all the point making the object is also transformed. There are four different transformations: Reflection, dilation, translation and rotation.

If y = f(x), y = f(x) + k is a translation k units up if k > 0 and is a translation k units down if k < 0

Given that the parent function is an absolute value represented by:

y = |x|

Therefore comparing y = |xl - 6 with y = f(x) + k, k < 0, therefore this is a translation of 6 units down

7 0

2 years ago