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murzikaleks [220]
3 years ago
12

Please help I’m stuck

Mathematics
2 answers:
solong [7]3 years ago
8 0

Answer:

23

hope it helps u :)

Step-by-step explanation:

2x+13

x=5

=2×5+13

=10+13

=23

anyanavicka [17]3 years ago
6 0

Answer:

23

Step-by-step explanation:

When I was first introduced to algebra, One thing that had me stumped for the longest time is terms like "2x".  This is because nobody explained to me that saying "2x" and similar terms implies a multiplication.  so "2x" actually means "2 × x".  I only mention this because I thought maybe it would help.

To solve it then, all we need to do then is think of "x" as meaning five, and it being multiplied by 2

We're given this:

2x + 13

which you can also read as this:

2 \times x + 13

We're told x equals five, so we can say:

2 \times 5 + 13

Remember when solving this, that multiplication and division <em>always</em> come before addition and subtraction, so let's do the multiplication:

10 + 13

and then add it up:

23

So the final answer is twenty-three.

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he zeros of a quadratic equation are the points where the graph of the quadratic equation crosses the x-axis.

Step-by-step explanation:

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Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
4 years ago
Help pls I’m in trouble I need this
nexus9112 [7]

Answer:

We conclude that:

  • g(-3) = 3.5

Step-by-step explanation:

Some background Concepts:

From the graph, it is clear that at x = -10, the graph intersects the x-axis.

So, the x-intercept of the graph is (-10, 0).

It means g(-10) = 0

From the graph, it is clear that at y = 5, the graph intersects the y-axis.

So, the y-intercept of the graph is (0, 5).

It means g(0) = 5

Determining g(-3):

From the graph, it is clear that at x = -3, the value of the function output = 3.5.

In other words,

at x = -3, g(-3) = 3.5

Therefore, we conclude that:

  • g(-3) = 3.5
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