Answer:
14. x=-2.5, y = -7
15. x=28 y = -20
Step-by-step explanation:
14. Let's solve this system by elimination. Multiply the first equation by -1.
-1*(4x-y)= -1*(- 3)
-4x +y = 3
Then add this to the second equation.
-4x+y = 3
6x-y= - 8
--------------
2x = -5
Divide each side by 2
x = -2.5
We still need to find y
-4x+y =3
-4(-2.5) + y =3
10 +y =3
Subtract 10 from each side.
y = 3-10
y = -7
15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11
11(
5x+6y)= 11*20
55x+66y = 220
Multiply the second equation by -6
-6(9x+11y)=32*(-6)
-54x-66y = -192
Add the modified equations together.
55x+66y = 220
-54x-66y = -192
---------------------------
x = 28
We still need to solve for y
5x+6y = 20
5*28 + 6y =20
140 + 6y = 20
Subtract 140 from each side
6y = -120
Divide by 6
y = -20
It's basically 5,000*2=10,000
and 10,000/2=5,000. your answer would be either they multiplied 5,000 by 2 and 10,000 divided by 2, they increased the sales by 5,000
By definition of absolute value, you have
or more simply,
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
Answer:
first blank 39, second 9/39 i think, and third 351
Step-by-step explanation:
i haven't done something like this in a long time so i dont know if its completely correct or correct at all
Answer:
7. r = -5
8. x = -1
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define</u>
r + 2 - 8r = -3 - 8r
<u>Step 2: Solve for </u><em><u>r</u></em>
- Combine like terms: -7r + 2 = -3 - 8r
- Add 8r to both sides: r + 2 = -3
- Subtract 2 on both sides: r = -5
<u>Step 3: Check</u>
<em>Plug in r into the original equation to verify it's a solution.</em>
- Substitute in <em>r</em>: -5 + 2 - 8(-5) = -3 - 8(-5)
- Multiply: -5 + 2 + 40 = -3 + 40
- Add: -3 + 40 = -3 + 40
- Add: 37 = 37
Here we see that 37 does indeed equal 37.
∴ r = -5 is a solution of the equation.
<u>Step 4: Define equation</u>
-4x = x + 5
<u>Step 5: Solve for </u><em><u>x</u></em>
- Subtract <em>x</em> on both sides: -5x = 5
- Divide -5 on both sides: x = -1
<u>Step 6: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: -4(-1) = -1 + 5
- Multiply: 4 = -1 + 5
- Add: 4 = 4
Here we see that 4 does indeed equal 4.
∴ x = -1 is a solution of the equation.
