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3 years ago

Graph the system. Tell whether the system has one solution, no solution, or infinitely many solutions. y = –2x + 1 y = –2x – 3

1 answer:

8 0

Let's try the actual solution of the system

y = –2x + 1
y = –2x – 3

Note that the slopes of the graphs of these two lines are the same: -2.

That means that the lines are parallel to one another.

Only the y-intercepts (1 and -3) are different.

Since the 2 lines never intersect, the system has no solution.

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I need help pls answer All 4 of the questions

Alja [10]

Answer:

Step-by-step explanation:

1: 7/6

2: 3/2

3: 0.000288

4: six tenths

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3 years ago

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Joaquin has $1,300 in the bank and has investments worth $4,000. He also has $7,000 worth of credit card debt. What is Joaquin's

svetlana [45]

The answer to this question would be: -$1,700

In this question, we can separate the account into positive/debit and negative/credit value. The positive/debit value should be:

Joaquin has $1,300 in the bank
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Total positive value = $1300 + $4000= $5300

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4 years ago

What is the distance between (-3, 4) and (4, 4)

svetoff [14.1K]

Answer:

7 units

Step-by-step explanation:

$\sqrt{|-3-4|^{2}-|4-4|^{2}}$

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3 years ago

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What is the percentage decrease if an item has dropped from $115 to $46

Sav [38]

Answer: 60% decrease

Step-by-step explanation:

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4 years ago

Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate y' terms: $-xy' - 2y' = y$
[Algebra] Factor y': $y'(-x - 2) = y$
[Algebra] Isolate y': $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

Step 3: Find y

Define equation: $-xy - 2y = -4$
Factor y: $y(-x - 2) = -4$
Isolate y: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

Step 4: Rewrite 1st Derivative

[Algebra] Substitute in y: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

Step 5: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$

Step 6: Find Slope at Given Point

[Algebra] Substitute in x: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

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3 years ago

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