At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount.
Let x be the amount of one of Derek's withdrawals
Each of John's withdrawals was $5 more than each withdrawal that Derek made.
x + 5 is t the amount of one of John's withdrawals
Derek made 7 withdrawals
So amount withdraw 7 times = 7x
John made 5 withdrawals
So amount withdraw 5 times = 5(x+5)
Both Derek and John withdrew the same amount of money in the end
(A) 7x = 5(x+5)
(B) Solve for x
7x = 5x + 25
Subtract 5x from both sides
2x = 25
Divide by 2
x = 12.5
(C) check your solution
we plug in 12.5 for x in 7x= 5x + 25
7(12.5) = 5(12.5) + 25
87.5 = 62.5+ 25
87.5 = 87.5
(D) Each brother withdrawal 87.5 dollars
X(x + 2) = 120sq units
<span>Set it equal to 0 </span>
<span>x^2 + 2x - 120 = 0 </span>
<span>factor </span>
<span>(x + 12)(x - 10) </span>
<span>For the shorter side: </span>
<span>x - 10 = 0 </span>
<span>x = 10 </span>
<span>Now that you have x, solve for the longer side which we said was represented by </span>
<span>x + 2 </span>
<span>10 + 2 = 12 </span>
<span>Proof </span>
<span>A = L x W </span>
<span>120 = 10 x 12 </span>
<span>120 = 120 </span>
<span>true </span>
<span>Our length is 12 and our width is 10</span>
Answer:
A
Step-by-step explanation:
6x-20= 118
6x= 118 + 20
6x = 138
x = 138/6
x= 23
Answer:
(2)
Step-by-step explanation:
One property of a parallelogram is
• opposite angles are congruent, hence
∠L = ∠N and ∠M = ∠P
Using ∠L = ∠N, then
8x - 41 = 6x + 3 ( subtract 6x from both sides )
2x - 41 = 3 ( add 41 to both sides )
2x = 44 ( divide both sides by 2 )
x = 22
-------------
∠M = 2x + 1 = (2 × 22) + 1 = 44 + 1 = 45°
∠P = ∠M = 45° → (2)
My apologies on answering late...
Same situation as the previous problem, but this time, all you need to do is state the degree of the angle instead of just providing the angle itself.
ΔABC ≅ ΔDEF
Now, we can see that ∠C ≅ ∠F. Using this information, we can find ∠C on the first triangle ( which is 
° ).
Since ∠C ≅ ∠F,
m∠F is °.
Hope I caught your question in time!
Have a good one! If you need anymore help, let me know.
