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Marat540 [252]
2 years ago
8

Please help me answer this.

Mathematics
1 answer:
TiliK225 [7]2 years ago
6 0
8x2+6x-2

I think! because 8,6, and 2 are all divisible by 2

and 3. Factor the quadratic

2(4x-1)(x+1)
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21 pointQuadrilateral QRST has diagonals that do NOT bisect each other, are congruent, and are not perpendicular. Which type of
Aleks [24]

Given:

Quadrilateral QRST has diagonals that do NOT bisect each other, are congruent and are not perpendicular.

By checking the options:

Rhombus, rectangle and the square each of them the diagonals bisect each other

So, the answer will be option 4) Isosceles Tra

8 0
1 year ago
Holly works at a garage.
Musya8 [376]
The answer is £287.98 is the correct answer
5 0
3 years ago
During the 2007 baseball season, Wade was up to bat 10 more times than
bearhunter [10]

Answer:

The number of times Ellis get to bat  is  558.

Step-by-step explanation:

Here, let us assume the number of times Ellis bat = m

So, the number of times Dwight bat =  17 times fewer than Ellis

                                                            = m - 17

Also, number of times Wade got to bat  = 10 more times than Dwight

= ( m- 17 )+ 10

Total number of bat times = 1650

So, the number of times ( Wade + Dwight + Ellis) bat together = 1650

⇒   ( m- 17 )+ 10 + (m - 17) + m =  1650

or, 3 m - 34 + 10 =  1650

or, 3  m = 1674

⇒ m = 1674/3 = 558 , or m = 558

Hence, the number of times Ellis get to bat = m = 558.

3 0
3 years ago
The equation of a circle is ​x2+4x+y2−10y+13=0​ .
kiruha [24]
X^2+4x+4+y^2-10y+25=-13+25+4

(x+2)^2 + (y-5)^2= 16
Third option
8 0
3 years ago
Read 2 more answers
How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com
Mnenie [13.5K]

<span>If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)


(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>

 

<span>If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)


(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>

 

<span>We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845</span>

Therefore, there are 4845 ways to have a group of 4 with 4 women.

 

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

 

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee


5 0
3 years ago
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