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Inga [223]
3 years ago
12

Can y’all please help ASAP

Mathematics
1 answer:
Leto [7]3 years ago
4 0
Y // -1 , 0 , 1 , 2 , 3 , 4 , 5
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Factor the trinomial: 3x2 + 11x + 6
OleMash [197]
3x^2+11x+6

(3x^2+2x) + (9x+6)

Factor out x  from 

3x^2+2x  =  x(3x+2)

Factor out 3  from

9x+6 = 3(3x+2)

= x(3x+2)+3(3x+2)

Factor out common  term (3x+2) :

=(3x+2)(x+3)

hope this helps!

5 0
3 years ago
HELP MEPLEASE IM DESPO
aliina [53]

Answer:

<h3>78.56%</h3>

Step-by-step explanation:

area of square: 16

area of circle: 12.57

12.57/16 = x/100

x=78.56

4 0
3 years ago
Which expressions represent Ernie's problem? Select each correct answer. 1500525 5251500 1500 ÷ 525 525 ÷ 1500 5251500 1500525
Kay [80]

Answer:

525 ÷ 1500 5251500

Step-by-step explanation:

3 0
3 years ago
Answer please i need
frosja888 [35]

Answer:

I'm pretty sure its C

Step-by-step explanation:

shhshshdhdhdjsjdncgc sg shhshshdhdhdjsjdncgc ghh Judy ch

5 0
3 years ago
A student walk 60m on a bearing
Katena32 [7]

Answer:

d = 234.6 m

Step-by-step explanation:

You can consider a system of coordinates with its origin at the beginning of the walk of the student.

When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:

x_1=60cos28\°=52.97m\\\\y_1=60sin28\°=28.16m

she is at the coordinates (52.97 , 28.16)m.

Next, when she walks 180m to the east, her coordinates are:

(52.97+180 , 28.16)m = (232.97 , 28.16)m

To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}\\\\x=232.97\\\\x_o=0\\\\y=28.16\\\\y_o=0\\\\d=\sqrt{(232.97-0)^2+(28.16-0)^2}m=234.6m

The displacement of the student on her complete trajectory was of 234.6m

8 0
3 years ago
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