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Furkat [3]
3 years ago
8

What is the perimeter of abc with vertices a(-2 9)?

Mathematics
1 answer:
vazorg [7]3 years ago
6 0
For finding the perimeter of abc (triangle), we MUST need the two others coordinates of b and c
let b(1, 2) and c (2, -4)
we need to calculate the distances ab, ac, and bc
vector (ab)=(1-(-2), 2-9)=(3, -7), so its length is ab=sqrt(3²+ (-7)²)=7.61
realising the same method, we find bc=6.08, ac=13.60

so the perimeter of abc is P= ab+bc+ca=13.60+6.08+7.61=27.29
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polet [3.4K]
You need to make both numbers the same denominator (bottom number). Do this by multiplying. So for the first question you need 15 as the denominator. 3x3=9 5x3=15 first one is 9/15. Then its 1x5=5 and 3x5=15. Then you add 9+5 to get 14. The denominator stays the same. 14/15. This cannot be reduced. If reduced, you would be able to divide both numbers by the same number.
1. 14/15
2. 5/6
3. 1/2
4. 13/24
5. 13/24
6. 9/10
7. 5/9
8. 11/35
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6 0
2 years ago
How many terms does the polynomial have? a 2 + b - cd 3 2 3 4
vaieri [72.5K]

ANSWER

3

EXPLANATION

The given polynomial is

{a}^{2}  + b - cd

The terms of the polynomial are:

First term:

{a}^{2}

Second term:

b

Third term:

- cd

Therefore the given polynomial has 3 terms.

7 0
3 years ago
Write an equation in slope-intercept form of the line shown. (-4,0) and (1,-5)
Irina18 [472]

The slope intercept form is y = -x - 4

To find the slope intercept form given a couple of points, start by finding the slope using the slope equation.

m(slope) = (y2 - y1)/(x2 - x1)

m = (-5 - 0)/(1 - -4)

m = -5/5

m = -1

Now we look for the intercept using slope intercept form, our slope and a point.

y = mx + b

0 = -4(-1) + b

0 = 4 + b

-4 = b

Now we can use those two things top model the equation.

y = -x - 4

5 0
3 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
True or false? "If two angles form a linear pair, they must be supplementary?"
vodka [1.7K]

Answer:

true?

Step-by-step explanation:

im guessing.

4 0
3 years ago
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