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Allisa [31]
3 years ago
10

PLEASE HELP!! 48 POINTS!!

Mathematics
1 answer:
Likurg_2 [28]3 years ago
4 0
The exact number is 121.966019, so if you round to the nearest whole number, the answer is 122.
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Six groups of students sell 162 balloons at the school carnival. There are 3 students in each group. If each student sells the s
choli [55]
162/6=27
27/3=9

9 balloons per student~!
4 0
4 years ago
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raj and his sister zia are both at secondary school. Raj is three years older than zia. the sum of the squares of their ages is
Cerrena [4.2K]

Let Zia's age be = x years

Raj is three years older than Zia, so Raj's age = x+3 years

As the sum of the squares of their ages is 369, so

x^{2}+(x+3)^{2}=369

x^{2}+x^{2}+6x+9=369

2x^{2}+6x+9=369

2x^{2}+6x-360=0

Factoring 2 out we have,

x^{2}+3x-180=0

Now solving this equation,

x^{2}+15x-12x-180=0

x(x+15)-12(x+15)

(x+15)(x-12)

x=-15 and x=12

Neglect the negative value as x cannot be negative

So x=12

Hence, Zia's age is 12 years and

Raj's age is 12+3 = 15 years


4 0
3 years ago
What would prove that the two triangles are<br> congruent?
aniked [119]
The lines on the sides!
5 0
3 years ago
Read 2 more answers
Can somebody help me with this?
lesya [120]

Answer:

141 °

since they are supplementary angles and they always add to 180 °

4 0
3 years ago
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An NFL coach sometimes uses a defense that utilizes 5 defensive linemen, 4 linebackers, and 2 defensive backs. His roster (the p
german

Answer:

there are 4725 different ways to pick the 11 players

Step-by-step explanation:

assuming that a player can only play from its position and not from the other 2 , then each defensive position is independent from the others , and since the order each player is chosen is not relevant we have that:

total combinations = possible combinations of linemen * possible combinations of linebackers * possible combinations of defensive backs = combinations of 5 linemen  from 7 * combinations of 4 linebackers  from 6 * combinations of 2 defensive backs from 6 =  [ 7!/(5!*(7-5)!] * [ 6!/(4!*(6-4)!] * [ 6!/(2!*(6-2)!] = 7!/(2!*5!)*[ 6!/(4!*2!)]² = 4725 combinations

thus there are 4725 different ways to pick the 11 players

8 0
3 years ago
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