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alexgriva [62]
4 years ago
14

Someone help please! k(-1)=______​

Mathematics
1 answer:
prohojiy [21]4 years ago
8 0

Answer:

<h2>k(-1) = 7</h2>

Step-by-step explanation:

Look at the picture.

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Llana [10]

Answer:

C. 12

Step-by-step explanation:

7 0
3 years ago
Slope of the line<br> (5,11) (-5,-1)
AnnZ [28]

Answer:

1.8

Step-by-step explanation:

To do this, you need the slope formula

y^2 - y^1   over

x^2  - x^1

x^1     y^2           x^2     x^1

( 5,     11)             (-5         -1)

-7 - 11  =  -18

-5 -5   =  -10   (divide)  = 1.8

4 0
4 years ago
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How many different three-letter initials are there that contain at least one a?
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I believe the answer is 600
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3 years ago
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Anyone want to help me with this? I will mark brainliest and thanks
Degger [83]

Answer:

x = -8

m∠BDC = 68°

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

Equality Properties

<u>Geometry</u>

  • Definition of a Line: A line is 180°

Step-by-step explanation:

<u>Step 1: Set up Equation</u>

m∠BDC + m∠CDA = 180°

(-7x + 12)° + (-8x + 48)° = 180°

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Combine like terms:                    -15x + 60 = 180
  2. Isolate <em>x</em> term:                              -15x = 120
  3. Isolate <em>x</em>:                                       x = -8

<u>Step 3: Find m∠BDC</u>

  1. Define:                    m∠BDC = (-7x + 12)°
  2. Substitute in <em>x</em>:       m∠BDC = (-7(-8) + 12)°
  3. Multiply:                  m∠BDC = (56 + 12)°
  4. Add:                        m∠BDC = 68°
6 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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