
Consider, LHS

We know,

We know,

So, using this identity, we get

can be rewritten as

<h2>Hence,</h2>


Let number of fives = x then number of ones = 3x and number of tens = x - 1
so we can create the equation
x + 3x + x-1 + y = 26 where y = number of twenties
so
5x + y = 27
also we have the equation
5x + 3x + 10(x - 1) + 20y = 120
18x + 20y = 130..................................(1)
5x + y = 27 multiply by -20:-
-100x - 20y = -540..............................(2)
Adding equation (1) and (2)
-82x = -410
x = 5, that is 5 fives
Now plug x = 5 into equation 1:-
18(5) + 20y = 130
20y = 40
y = 2 , that is 2 twenties
So the answer is there are (3x) = 15 ones , 5 fives, 4 tens and 2 twenties
Answer:
-37
Step-by-step explanation:
Answer:
D.Lindsay makes 480 candles. She divides 480 by 15 to get 32. She does not make enough candles.
Step-by-step explanation:
The explanation of the sufficient candles that are required to check whether she reach her goal or not is as follows:
She makes 480 candles that come from
= 24 candles each days × 20 days
= 480 candles
Now if we divide the 480 candles from 15 so we get 32
This is less than the minimum candles required i.e. 35
So, the option D is correct
Answer:
There are 30 students in the class
Step-by-step explanation:
we can write a ratio as:
8:10 as 24:x
Writing this as an equation gives:
8
10
=
24
x
We can now solve for
x
while keeping the equation balanced:
10
x
×
8
10
=
10
x
×
24
x
10
x
×
8
10
=
10
x
×
24
x
x
×
8
=
10
×
24
8
x
=
240
8
x
8
=
240
8
8
x
8
=
30
x = 30