Answer:
the first one because supplementary means angles which add up to 180 and 180 degrees is a straight line and the third one because they both share the same vertex and side
Step-by-step explanation:
Answer:
248
Step-by-step explanation:
Solution for What is 400 percent of 62:
400 percent *62 =
(400:100)*62 =
(400*62):100 =
24800:100 = 248
Now we have: 400 percent of 62 = 248
Question: What is 400 percent of 62?
Percentage solution with steps:
Step 1: Our output value is 62.
Step 2: We represent the unknown value with $x$.
Step 3: From step 1 above,$62=100\%.
Step 4: Similarly, x=400\%.
Step 5: This results in a pair of simple equations:
62=100\%(1).
x=400\%(2).
Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both
equations have the same unit (%); we have
\frac{62}{x}=\frac{100\%}{400\%}
Step 7: Again, the reciprocal of both sides gives
\frac{x}{62}=\frac{400}{100}
\Rightarrow x=248
Therefore, 400 of 62 is 248
The answer is: " 91 " .
___________________________________________________
→ " B = 91 " .
__________________________________________________
Explanation:
__________________________________________________
Given:
__________________________________________________
" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
_____________________________________________________
METHOD 1)
_____________________________________________________
Solve for "x" ; and then plug the solved value for "x" into the expression given for "B" ; to solve for "B"
_____________________________________________________
(115 − 2x) + (169 − 6x) =
115 − 2x + 169 − 6x = ?
→ Combine the "like terms" ; as follows:
+ 115 + 169 = + 284 ;
− 2x − 6x = − 8x ;
_________________________________________________________
And rewrite as:
" − 8x + 284 " ;
_________________________________________________________
→ " - 8x + 284 = 180 " ;
Subtract: "284" from each side of the equation:
→ " - 8x + 284 − 284 = 180 − 284 " ;
to get:
→ " -8x = -104 ;
Divide EACH SIDE of the equation by "-8 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104/-8 ;
→ x = 13
__________________________________________________________
Now, to find the value of "B" :
__________________________________________________________
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
↔ B = 169 − 6x ;
= 169 − 6(13) ; ===========> Plug in our "solved value, "13", for "x" ;
= 169 − (78) ;
= 91 ;
B = " 91 " .
__________________________________________________
The answer is: " 91 " .
____________________________________________________
→ " B = 91 " .
____________________________________________________
Now; let us check our answer:
____________________________________________________
→ A + B = 180 ;
____________________________________________________
Plug in our "solved answer" ; which is "91", for "B" ; as follows:
________________________________________________________
→ A + 91 = ? 180? ;
↔ A = ? 180 − 91 ? ;
→ A = ? -89 ? Yes!
________________________________________________________
→ " A = -2x + 115 " ; ↔ A = 115 − 2x ;
Plug in our solved value for "x"; which is: "13" ;
" A = 115 − 2x " ;
→ A = ? 115 − 2(13) ? ;
→ A = ? 115 − (26) ? ;
→ A = ? 29 ? Yes!
_________________________________________________
METHOD 2)
_________________________________________________
Given:
__________________________________________________
" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ Solve for the value of "B" :
_______________________________________________________
A + B = 180 ;
→ B = 180 − A ;
→ B = 180 − (115 − 2x) ;
→ B = 180 − 1(115 − 2x) ; ==========> {Note the "implied value of "1" } ;
__________________________________________________________
Note the "distributive property" of multiplication:__________________________________________________ a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b − c) = ab − ac .________________________________________________________
Let us examine the following part of the problem:
________________________________________________________
→ " − 1(115 − 2x) " ;
________________________________________________________
→ " − 1(115 − 2x) " = (-1 * 115) − (-1 * 2x) ;
= -115 − (-2x) ;
= -115 + 2x ;
________________________________________________________
So we can bring down the: " {"B = 180 " ...}" portion ;
→and rewrite:
_____________________________________________________
→ B = 180 − 115 + 2x ;
→ B = 65 + 2x ;
_____________________________________________________
Now; given: "B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ " B = 169 − 6x = 65 + 2x " ;
______________________________________________________
→ " 169 − 6x = 65 + 2x "
Subtract "65" from each side of the equation; & Subtract "2x" from each side of the equation:
→ 169 − 6x − 65 − 2x = 65 + 2x − 65 − 2x ;
to get:
→ " - 8x + 104 = 0 " ;
Subtract "104" from each side of the equation:
→ " - 8x + 104 − 104 = 0 − 104 " ;
to get:
→ " - 8x = - 104 ;
Divide each side of the equation by "-8" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104 / -8 ;
to get:
→ x = 13 ;
______________________________________________________
Now, let us solve for: " B " ; → {for which this very question/problem asks!} ;
→ B = 65 + 2x ;
Plug in our solved value, " 13 ", for "x" ;
→ B = 65 + 2(13) ;
= 65 + (26) ;
→ B = " 91 " .
_______________________________________________________
Also, check our answer:
_______________________________________________________
Given: "B = - 6x + 169 " ; ↔ B = 169 − 6x = 91 ;
When "x = 13 " ; does: " B = 91 " ?
→ Plug in our "solved value" of " 13 " for "x" ;
→ to see if: "B = 91" ; (when "x = 13") ;
→ B = 169 − 6x ;
= 169 − 6(13) ;
= 169 − (78)______________________________________________________
→ B = " 91 " .
______________________________________________________
Answer:
Step-by-step explanation:
Let 
y=2x-10
Switch x and y,
x=2y-10
x+10=2y
(x/2)+5=y
So 
Given:
The equation is,

Explanation:
Simplify the equation by using logarthimic property.

Simplify further.

Solve the quadratic equation for x.

From the above equation (x - 6) = 0 or (x - 3) = 0.
For (x - 6) = 0,

For (x - 3) = 0,

The values of x from solving the equations are x = 3 and x = 6.
Substitute the values of x in the equation to check answers are valid or not.
For x = 3,

Equation satisfy for x = 3. So x = 3 is valid value of x.
For x = 6,

Equation satifies for x = 6.
Thus values of x for equation are x = 3 and x = 6.