Answer:
- Solution of equation ( q ) = <u>1</u><u>6</u>
Step-by-step explanation:
In this question we have given an equation that is <u>3 </u><u>(</u><u> </u><u>q </u><u>-</u><u> </u><u>7</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u>and we have asked to solve this equation that means to find the value of <u> </u><u>q</u><u> </u><u>.</u>
<u>Solution : -</u>

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis :

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 21 on both sides :

On further calculations we get :

<u>Step </u><u>3 </u><u>:</u> Dividing by 3 from both sides :

On further calculations we get :

- <u>Therefore</u><u>,</u><u> </u><u>solution</u><u> </u><u>of </u><u>equation</u><u> </u><u>(</u><u> </u><u>q </u><u>)</u><u> </u><u>is </u><u>1</u><u>6</u><u> </u><u>.</u>
<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>
Now we are very our answer by substituting value of q in the given equation . So ,
<u>Therefore</u><u>,</u><u> </u><u>our </u><u>solution</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u>
<h2>
<u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
Answer:
D 7
Step-by-step explanation:
Total no. of bulb= 5
no. of defected bulb= 3
no. of not defected bulb=2
Total no. of bulb combination = 5C2
=5!/2!(5-2)!
= 5!/2!3!
= 5×4×3×2×1/2×1×3×2×1
=120/12
=10
( since a room can be lighted with one bulb also)
total no. of bulb combination when room shall not light = 3C2
3!/2!(3-2)!
= 3!/2! 1!
= 3×2×1/2×1×1
= 6/2
=3
Now,
Total no. of trial when room shall light
=10-3
=7
Hence, number of trial when the room shall be lighted is 7 which is option d
Answer:
y= -31
Step-by-step explanation:
8.5(-4)= -34
-34+3= -31
Answer:
the answer is 3/5
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