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3 years ago

Solve the following: -x + 4(1 + 5x) = 156 I need help please ?

Mathematics

1 answer:

Elena L [17]3 years ago

3 0

I believe the answer is x=8

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Which graph represents a quadratic function with a vertex at (0, 0)

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PARABOLA

On a coordinate plane, a parabola opens up. It goes through (negative 5, 6), has a vertex of (0, 0), and goes through (5, 6).

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3 years ago

A number decreased by 16 is -26. Find the number

lidiya [134]

-42 decreased by 16 = -26

4 0

3 years ago

Identify the vertex for the graph of y = 2x2 + 8x − 3. (2, 21) (2, 17) (−2, −11) (−2, −27)

Finger [1]

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{lcccl}
y = & 2x^2& +8x& -3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad 
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{8}{2(2)}~~,~~-3-\cfrac{8^2}{4(2)} \right)\implies \left( \cfrac{-8}{4}~~,~~-3-\cfrac{64}{8} \right)
\\\\\\
\left(-2~~,~~-3-8 \right)\implies (-2~~,~~-11)$

5 0

3 years ago

Read 2 more answers

Represent the geometric series using the explicit formula.

Dennis_Churaev [7]

The correct answer is:

b.) f(n) = 3 • (-2)^(n-1)

Further explanation:

Given sequence is:

3, -6, 12, -24, ...

We have to find the common ratio first.

Common ratio is the ratio between two consecutive terms of a geometric sequence.

It is denoted by r.

So,

$Here,\\a_1=3\\a_2=-6\\a_3=12\\a_4=-24\\r=\frac{a_2}{a_1}=\frac{-6}{3}=-2\\\frac{a_3}{a_2}=\frac{12}{-6}=-2$

General formula for geometric sequence is:

$a_n=a_1r^{n-1}\\Here\\a_n\ is\ the\ nth\ term\\a_1\ is\ the\ first\ term\\and\\r\ is\ common\ ratio$

Putting the values of a1 and r

$a_n=3.(-2)^{n-1}\\f(n)=3.(-2)^{n-1}$

Hence,

The correct answer is:

b.) f(n) = 3 • (-2)^(n-1)

Keywords: Geometric Sequence, Explicit formula

Learn more about geometric sequence at:

#LearnwithBrainly

7 0

3 years ago

What is the difference in simplest form? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bn%5E2%2B3n%2B2%7D%7Bn%5E2%2B6n%2B8%7D-%5

Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$

3 0

3 years ago