No solution is the correct answer i think
Answer:
hold up im solving it
Step-by-step explanation:
Answer:
The solution of the linear equation is: (6,7)
Step-by-step explanation:
This being that:
x + y = 13
1/2x + y = 10
What you do in order to find the solution is either (2 ways):
- Substitution
- Elimination
I'm sure that either way, you'll get the same thing. I used: Substitution.
To do this choose either if you want to find [x] or [y] first, then substitute it in. I will choose to find [x] first.
<h2>x + y = 13</h2><h2><u> - y </u> </h2><h2>x = 13 - y</h2>
Now substitute it into:
1/2x + y = 10
1/2(13 - y) + y = 10
1/2(13 -y) + y= 10
<h2>Then to get rid of fraction, instead of dividing, you multiply with reciprocal</h2><h2>1/2 · 2/1 (They cancel out) Leaving you with:</h2><h3 /><h3>(13 - y) 2(y) = 2(10)</h3><h3>(Everything is multiplied, except for 13 - y, because it had parenthesis to protect it.)</h3><h3>13 - y + 2y = 20</h3><h3 /><h2>13 + y = 20</h2><h2><u>- 13 - 13</u></h2><h2> <u>y = 7</u></h2><h2 /><h2>You now have your [y] for your coordinate.</h2><h2>(x, 7), now time to find your [x]</h2><h2 />
Now you substitute your [y] into your equation:
1/2x + (7) = 10
<h2>
1/2x + 7 = 10</h2><h2><u>
-7 = -7</u></h2><h2>
1/2x = 3</h2><h2>
(To get rid of fraction, multiply on both sides with reciprocal)</h2><h2>
1/2 cancels out with 2/1</h2><h2>
x = 2(3)</h2><h2>
x = 6</h2><h2>You now have your [x] coordinate.</h2><h2 /><h2>This is your coordinate: </h2><h2>(6,7)</h2><h2>
</h2>
Answer:
$1876.31
Step-by-step explanation:
The present value can be found using the compound interest formula.
<h3>Present value</h3>
The formula for the future value of an investment of P earning annual interest rate r compounded n times per year for t years is ...
FV = P(1 +r/n)^(nt)
Filling in the known values gives ...
4000 = P(1 +0.085/4)^(4·9) = P(1.02125^36)
Then the amount to be invested is ...
P = 4000/1.02125^36 ≈ 1876.31
The present value is $1876.31.
Answer:
- x = 30°
- DB = 26
- AD = BC = AB = DC = 7
Step-by-step explanation:
- <em>Diagonals of a square are congruent and perpendicular and bisect each other</em>
<h3>Q4</h3>
m∠AEB = 3x
m∠AEB = 90°
<h3>Q5</h3>
AE = 3x - 2
EC = 2x + 3
- AE = EC
- 3x - 2 = 2x + 3
- 3x - 2x = 3 + 2
- x = 5
DB = EC = 2(AE) = 2(3*5 - 2) = 2(13) = 26
<h3>Q6</h3>
<u>AD and BC are the sides, which are equal</u>
- 2x - 1 = 5x - 13
- 5x - 2x = 13 - 1
- 3x = 12
- x = 4
AD = BC = AB = DC = 2*4 - 1 = 7
