Answer:
x = 2
Step-by-step explanation:
You are solving for the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.
PEMDAS is the order of operation, and =
Parenthesis
Exponents (& Roots)
Multiplication
Division
Addition
Subtraction.
First, subtract 25 from both sides of the equation:
5x + 25 (-25) = 35 (-25)
5x = 35 - 25
5x = 10
Isolate the variable, x. Divide 5 from both sides:
(5x)/5 = (10)/5
x = 10/5
x = 2
x = 2 is your answer.
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Answer:
20 = 2 + (7 -4) × 6
Step-by-step explanation:
The Order of Operations requires the parentheses be evaluated first, then the multiplication performed. Finally, the addition is performed.
If each of the blanks is filled with a single digit, the result of the multiplication must be a composite number greater than 10. Those are 12, 14, 15, 16, 18, 20. For the expression shown above, we have chosen to make the product be 18. That means the first blank is filled with 2 and the remaining blanks must evaluate to one of the products 2×9 or 3×6.
We have chosen 6 for the last blank, so the two blanks in parentheses must have a difference of 3. The digits 2 and 6 cannot be used, leaving possible choices as (3-0), (4-1), (7-4), (8-5).
Our final expression is chosen to be ...
20 = 2 +(7 -4)×6
Point I think is the anwser
Answer:
yea ur correct
Step-by-step explanation:
use PhotoMath or something like that to check next time, it'll save a lot of time
Answer:
Step-by-step explanation:
The Side-Angle-Side method cana only be used when information given shows that an included angle which is between two sides of a ∆, as well as the two sides of the ∆ are congruent to the included side and two sides of the other ∆.
Thus, since John already knows that 
and 
, therefore, an additional information showing that the angle between 
and 
in ∆ABC is congruent to the angle between 
and 
in ∆DEF.
For John to prove that ∆ABC is congruent to ∆DEF using the Side-Angle-Side method, the additional information needed would be .
See attachment for the diagram that has been drawn with the necessary information needed for John to prove that ∆ABC is congruent to ∆DEF.
