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topjm [15]
3 years ago
12

FUNCTIONS: In the space provided, type the answer in descending order as it applies without any spaces between the letters, numb

ers, or symbols.
Type the composition (fog)(x) of the given functions:
f(x) = x^2 + 2x − 6 and g(x) = x + 5.
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

Hence The composition (fog)(x) of the given function is x^2+12x+29.

Step-by-step explanation:

Given:

f(x) = x^2+2x-6

g(x)=x+5

We need to find (f o g)(x).

Solution:

Now we can say that;

(f o g)(x) = f(g(x))

(fog)(x) = (x+5)^2+2(x+5)-6

Now Applying distributive property we get;

(fog)(x) = (x+5)^2+2\times x+2\times5-6\\\\(fog)(x) = (x+5)^2+2x+10-6\\\\(fog)(x) = (x+5)^2+2x+4

Now Solving the exponent function we get;

(fog)(x) = x^2+2\times x\times 5+5^2+2x+4\\\\(fog)(x) = x^2+10x+25+2x+4\\\\(fog)(x) = x^2+12x+29

Hence The composition (fog)(x) of the given function is x^2+12x+29.

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Answer:

10

Step-by-step explanation:

4 0
3 years ago
Which is the standard form of the equation of the parabola that has a vertex of (3, 1) and a directric of x = -2?
vagabundo [1.1K]
Check the picture below.  So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\

\bf -------------------------------\\\\
\begin{cases}
h=3\\
k=1\\
p=5
\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)
\\\\\\
\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}

8 0
3 years ago
Read 2 more answers
An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam
Genrish500 [490]

Answer:

<h2>1.84feet</h2>

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range R = U\sqrt{\frac{H}{g} }

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}

Squaring both sides we have;

(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} =  \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet

The arrow will hit the target 1.84feet below the bullseye.

5 0
3 years ago
If a restaurant bills totals $85, what amount should be left as a 15$ tips
IgorC [24]
If the total bill is $85, and you want to leave a 15% tip, then you can calculate it as follows:

85 * 15%

= 85 * 15/100

= 85 * 0.15

= 12.75

The amount that should be left as a tip is D. $12.75.
6 0
3 years ago
The table summarizes the result of spinning the spinner shown.
weqwewe [10]
Hello!

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Our answer is B) 1/3.

Hope this helped!
5 0
3 years ago
Read 2 more answers
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