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Gelneren [198K]
3 years ago
6

Pls help me with these 2 math question will mark brainiest.

Mathematics
2 answers:
nika2105 [10]3 years ago
8 0

Answer:

3) B

4) A

Step-by-step explanation:

3)

<u>Formula</u>: y = mx + b

<u>Slope</u>: \frac{-3 - 1}{-4 - (-2)} =2

<em>Next, I substituted the numbers from both the chart and the slope.</em>

y=2x+b

(x, y) --> (1, 7)

\\7=2(1)+b\\

7=2+b

b=5\\

<em>Your final equation should be, </em>y=2x+5.

4)

<em>From the graph you can see (0, 4) is the y-intercept and the spaces are moving up twice and once to the right, making the slope 2. </em>

<em>Your answer should be, </em>y=2x+4.

zalisa [80]3 years ago
6 0
I’m not sure what the first one is, but number 4 is A) y=2x+4
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Graph a line with a slope of -5 that contains the point -3,-4
guapka [62]

Answer:

The equation of the line is 5x + y + 19 = 0

Step-by-step explanation:

The equation of the line with slope 'm' and given a point (x₁, y₁) passing through it we use the Slope - one - point form which is given by:

                                    y - y₁ = m(x - x₁)

The point given is: (-3, -4) and the slope is -5.

We get the equation of the line to be:

y - (-4) = -5(x - (-3))

⇒ y + 4 = -5(x + 3)

⇒ y + 4 = -5x - 15

⇒ 5x + y + 19 = 0. is the required equation of the line.

4 0
3 years ago
Which linear system of equations does the matrix represent 4 -7 -14 4 -12 4
Novosadov [1.4K]
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   4x  - 7y = -12
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Good luck!!!
4 0
3 years ago
How many years will $400 yield an interest of $112 at 14% simple interest
german

Answer:

2 years

Step-by-step explanation:

$400 x 14% = $56

$56 x 2 = $112

3 0
3 years ago
If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form
algol13

Answer:

\frac{d^2y}{dx^2} = \frac{-4}{3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

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:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Basic Power Rule]:                                                  -y'-6x^2=2yy'
  2. [Algebra] Isolate <em>y'</em> terms:                                                                              -6x^2=2yy'+y'
  3. [Algebra] Factor <em>y'</em>:                                                                                       -6x^2=y'(2y+1)
  4. [Algebra] Isolate <em>y'</em>:                                                                                         \frac{-6x^2}{(2y+1)}=y'
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-6x^2}{(2y+1)}

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}
  2. [Derivative] Simplify:                                                                                       y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}
  3. [Derivative] Back-Substitute <em>y'</em>:                                                                     y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}
  4. [Derivative] Simplify:                                                                                      y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}

<u>Step 4: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em> and <em>y</em>:                                                                     y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}
  2. [Pre-Algebra] Exponents:                                                                                      y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}
  3. [Pre-Algebra] Multiply:                                                                                   y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}
  4. [Pre-Algebra] Add:                                                                                         y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}
  5. [Pre-Algebra] Exponents:                                                                               y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}
  6. [Pre-Algebra] Divide:                                                                                      y''(-1,-2) = \frac{-36+24 }{9}
  7. [Pre-Algebra] Add:                                                                                          y''(-1,-2) = \frac{-12}{9}
  8. [Pre-Algebra] Simplify:                                                                                    y''(-1,-2) = \frac{-4}{3}
6 0
3 years ago
What is the domain of ƒ(x) if ƒ(x) = 1 x – 4?
Alina [70]
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