Can someone please help me on this question

Mathematics

1 answer:

Step2247 [10]3 years ago

6 0

Answer:

https://www.tiger-algebra.com/drill/4(2%e2%80%93x)%3E%e2%80%932x%e2%80%933(4x_1)/

Step-by-step explanation:

just put it in it gives you the answer

You might be interested in

Can you please help me with this i will give you a Brainiest

il63 [147K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the 52nd term of the sequence below?

VladimirAG [237]

Because the difference between any term and the previous term is a constant, this is an arithmetic sequence because of that constant which is referred to as the common difference, d. Which in this case is -35--38=-32--35=3

Any arithmetic sequence can be expressed as:

a(n)=a+d(n-1), a=initial term, d=common difference, n=term number.

In this case a=-38 and d=3 so

a(n)=-38+3(n-1) which we can simplify to

a(n)=-38+3n-3

a(n)=3n-41, so the 52nd term is:

a(52)=3(52)-41

a(52)=156-41

a(52)=115

3 0

3 years ago

The North Star is approximately 4,085 trillion kilometers away from earth. If 1 light

kumpel [21]

Answer:

I don't get the question meaning

6 0

3 years ago

A teacher asks students to list the math lessons they learned on nine random school days in the past month. She evaluates the st

Fittoniya [83]

Simple random sampling.

4 0

3 years ago

Read 2 more answers

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

3 years ago

Read 2 more answers