This is how you solve it:
3 + j / 11 = 2
3 + j = 2 × 11
3 + j = 22
j = 22 - 3
j = 19
Now you could check your answer by substituting the value of j :
3 + j / 11 = 2
3 + 19 / 11 = 2
22/11 = 2
2 = 2
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
t/15-t/35-(1/15)=0
4.1 Find the Least Common Multiple
The left denominator is : 15
The right denominator is : 35
Least Common Multiple:
105
4.2 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M = L.C.M / L_Deno = 7
Right_M = L.C.M / R_Deno = 3
4.3 Rewrite the two fractions into equivalent fractions
Two fractions are called equivalent if they have the same numeric value.
For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.
To calculate equivalent fraction , multiply the Numerator of each fraction, by its respectiveMultiplier.
4.4 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
5.1 Find the Least Common Multiple
The left denominator is : 105
The right denominator is : 15
Least Common Multiple:
105
5.2 Calculate multipliers for the two fractions
Denote the Least Common Multiple by L.C.M
Denote the Left Multiplier by Left_M
Denote the Right Multiplier by Right_M
Denote the Left Deniminator by L_Deno
Denote the Right Multiplier by R_Deno
Left_M = L.C.M / L_Deno = 1
Right_M = L.C.M / R_Deno = 7
5.3 Rewrite the two fractions into equivalent fractions
5.4 Adding up the two equivalent fractions
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
Now, on the left hand side, the 105 cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
4t-7 = 0
6.2 Solve : 4t-7 = 0
Add 7 to both sides of the equation :
4t = 7
Divide both sides of the equation by 4:
t = 7/4 = 1.750
A. What is the range?
B. What does the range mean in terms of the problem?
A. When is the function decreasing?
B. What does this decreasing interval mean in terms of the problem?
Answer:
Step-by-step explanation:
Q.No.5
A) Domain is all the values that x can take. In the graph we see that domain is
the values of x from 0 to 12 hours
Domain is 0≤x≤12
B) Domain is the no of hours as per the graph i.e. lower limit 0 to 12 hours
A) Range is the values what y can take. Here the values temperature can take
From the graph we find that range is -2 to 4
-2≤y≤4
B) Range is the values what y can take. Here the values temperature can take
A) Decreases whenever y decreases as x increases. Here. decreases for
x=0 to x=2
B) Decreasing interval is (0,2) and this means whenever number of hours increases the temperature decreases.
Answer:
5
Step-by-step explanation:
Answer:
TU= 8
WU=10
TX=5
TV=10
Step-by-step explanation: