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alexdok [17]
3 years ago
15

Using the graph, determine the coordinates of the roots of the parabola.

Mathematics
1 answer:
Liula [17]3 years ago
4 0

Answer:

(7, 0) and (3,0)

Step-by-step explanation:

On the graph, the parabola crosses the x-axis at two points. The two points are called the roots. And since they cross at (7, 0) and (3, 0), they are our roots. Hope this helps!

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Help!!I don't know how to answer this which one is it ???
7nadin3 [17]

Answer:

- 8 \frac{7}{12}

explanation:

y + 12 \frac{1}{3}  = 3 \frac{3}{4}  \\ subtract \: 12 \frac{1}{3}  \: from \: both \: sides \\  \\ y =  - 8 \frac{7}{12}

5 0
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Cassie has 10 feet of string to make 15 beaded bracelets. She needs 9 inches
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3 years ago
HELP GEOMOTRY QUESTION
V125BC [204]

Answer:

d) The points on the perpendicular bisector of a side of a triangle are equidistant from the  vertices of the side it bisects.

Step-by-step explanation:

By the property of perpendicular bisector, any point on the perpendicular bisector is equidistant from the end points of the segment it bisects.

For the given triangle any point on the line CD is equidistant from the vertices A and B as line CD  is the perpendicular bisector of side AB .

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d) The points on the perpendicular bisector of a side of a triangle are equidistant from the  vertices of the side it bisects.

4 0
3 years ago
A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, ev
Olin [163]

Answer:

The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Step-by-step explanation:

Given : A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat).

To find : The probability that no one sits in the same seat on both days of that week ?

Solution :

Let A_i be the i-th student sits on seat which he has been sitting on Monday.

According to question,

We have to calculate P(\cap^{20}_{i=1}A_i^c)

Applying inclusion exclusion formula,

P(\cap^{20}_{i=1}A_i^c)=1-P(\cap^{4}_{i=1}A_i)

P(\cap^{20}_{i=1}A_i^c)=1-P(A_1)+...+P(A_{20})-P(A_1\cap A_2)+...+P(A_{19}\cap A_{20})+P(A_1\cap A_2\cap A_3)+...+P(A_{18}\cap A_{19}\cap A_{20})....-P(A_1\cap A_2...\cap A_{20})

Using symmetry,

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}P(A_1\cap ...\cap A_k)

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}\frac{(20-k)!}{20!}

P(\cap^{20}_{i=1}A_i^c)=1+\sum^{20}_{k=1}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\sum^{20}_{k=0}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Therefore, The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

8 0
3 years ago
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