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3 years ago

Find the value of x I don’t understand this

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

101

Step-by-step explanation:

Pythagorean Theorem:

$A^{2}$ + $B^{2}$ = $C^{2}$

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F(x) = 3x-1 , (f of g)(x) = x^2 , Find g(x)

fomenos

Answer:

$g(x)=\dfrac{x^2+1}{3}$

Step-by-step explanation:

Given functions:

$\begin{cases}f(x)=3x-1\\(f \circ g)(x)=x^2\end{cases}$

Composite functions are when the output of one function is used as the input of another.

Therefore, the given composite function (f o g)(x) means to substitute function g(x) in place of the x in function f(x).

$\begin{aligned}(f \circ g)(x) & = x^2\\f(g(x)) & = x^2\\\implies 3(g(x))-1 & = x^2\\3(g(x))-1+1 & = x^2+1\\3(g(x))&=x^2+1\\\dfrac{3(g(x))}{3}&=\dfrac{x^2+1}{3}\\\implies g(x)&=\dfrac{x^2+1}{3}\end{aligned}$

Therefore:

$\boxed{g(x)=\dfrac{x^2+1}{3}}$

6 0

1 year ago

PLEASE HURRY, WILL GIVE BRAINIEST

Dafna11 [192]

There are 7 fruit bars left from each box

5 0

3 years ago

What are the exact values of sin theta cos theta tan theta if (3,-4) is a point on the terminal side of theta?

malfutka [58]

$\bf (\stackrel{a}{3}~,~\stackrel{b}{-4})\impliedby \textit{now let's find the \underline{hypotenuse}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{3^2+(-4)^2}\implies c=\sqrt{25}\implies \boxed{c=5}$

$\bf -------------------------------\\\\
sin(\theta )=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\qquad cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\qquad tan(\theta )=\cfrac{\stackrel{opposite}{-4}}{\stackrel{adjacent}{3}}$

6 0

3 years ago

Find the y value for point I such that DI and EI form a 3:5 ratio.

Pachacha [2.7K]

The y value of I is (5(2) + 3(5))/(3 + 5) = (10 + 15)/8 = 25/8 = 3.125

3 0

3 years ago

Hi Help me with question 6b) pls

Romashka-Z-Leto [24]

Answer:

k = 9

Step-by-step explanation:

Consider the pattern on the right side , that is

$T_{k}$ = k(k + 2) + 1 = k² + 2k + 1

Equating

k² + 2k + 1 = 100 ( subtract 100 from both sides )

k² + 2k - 99 = 0 ← in standard form

(k + 11)(k - 9) = 0 ← in factored form

Equate each factor to zero and solve for k

k + 11 = 0 ⇒ k = - 11

k - 9 = 0 ⇒ k = 9

But k > 0 , then k = 9

5 0

2 years ago