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Trava [24]
3 years ago
5

Find f(g(x)) & g(f(x)), if f(x) = (2x+1)^1/2 & g(x) = x^2 + 1.

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:b

Step-by-step explanation:

swat323 years ago
5 0

Answer:

is this a real problem ahdkjshfjkhdfw

Step-by-step explanation:

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Help!!!!Find the slope of the line on the graph.
artcher [175]

Answer:

m=\frac{-1}{2}

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

  • Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Find points from graph.</em>

Point (0, -4)

Point (-8, 0)

<u>Step 2: Find slope </u><em><u>m</u></em>

  1. Substitute:                    m=\frac{0+4}{-8-0}
  2. Add/Subtract:               m=\frac{4}{-8}
  3. Simplify:                        m=\frac{-1}{2}
8 0
3 years ago
Leah is creating satin bows for a wedding. From a 15-meter-long ribbon, she first cuts a piece 3 4/9 meters long. Then she cuts
grandymaker [24]
C is the answer. Hopes that helps.
8 0
3 years ago
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Which two grids can be combined to represent 140%
Brums [2.3K]

Answer:

C and D

Step-by-step explanation:

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3 years ago
George's page contains twice as many typed words as bills page and bills page contains 50 fewer words than Charlie's page. If ea
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Bill inititally had 110 words

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3 years ago
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Find the equation of the parabola with focus (5, 1) and directrix y = -1.
Mumz [18]
Check the picture below.

since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".

keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.

since the parabola is opening upwards, the value for "p" is positive, thus

\bf \textit{parabola vertex form with focus point distance}&#10;\\\\&#10;\begin{array}{llll}&#10;4p(x- h)=(y- k)^2&#10;\\\\&#10;\boxed{4p(y- k)=(x- h)^2}&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ( h, k)\\\\&#10; p=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;\begin{cases}&#10;h=5\\&#10;k=0\\&#10;p=1&#10;\end{cases}\implies 4(1)(y-0)=(x-5)^2&#10;\\\\\\&#10;4y=(x-5)^2\implies  y=\cfrac{1}{4}(x-5)^2

8 0
3 years ago
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