All categories

3 years ago

Can you help thanks this is my last one!

Mathematics

1 answer:

meriva3 years ago

4 0

Each box should be 2. If you subtract the number next to x on both sides of each inequality, you end up with x > 2.

PS. Thanks for marking me brainliest on the other problem :)

You might be interested in

How do i solve this problem 3x-y+z=-1 2x+3y+z=4 5x+4y+2z=5

Nonamiya [84]

Answer: use the order of operations and combine like terms

Step-by-step explanation:

what you do to one side of the equals sign you must do to the other

3 0

3 years ago

1 ⩽x ⩽ 5, for √(x³ + 36 )

alexandr402 [8]

$\bf ~\hspace{10em}1\le x\le 5~\hspace{5em}\sqrt{x^3+36}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\sqrt{x^3+36}\implies \sqrt{x^{2+1}6^2}\implies \sqrt{x^2x6^2}\implies \sqrt{(6x)^2x}\implies 6x\sqrt{x}
\\\\\\
\stackrel{\textit{x = 4}}{6(4)\sqrt{4}}\implies 24\sqrt{2^2}\implies 24\cdot 2\implies 48$

that's how I read it.... to get some value between 1 and 5, namely 4, to make the expression a rational, well, 48 can be expressed as 48/1.

7 0

3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago

How do we find the more precise measurement of 1 ft. ; 12 in.?

MariettaO [177]

You could do it in centimeters, which would be ABOUT 30.5 centimeters in one foot

8 0

3 years ago

Suppose you set up a function to show how many hot dogs you will purchase for a dinner when you have already bought two packages

bezimeni [28]

The domain of the function is:

h > 0, only integers.

Then the correct option is the first one.

<h3>What is the domain of the given function?</h3>

For a function f(x), we define the domain as the set of the possible values of x that we can use as inputs in the given function.

Here the function is:

f(h) = 6*h + 12

Where h is the number of packages that you buy.

Then h can be only integers larger than zero (as you can't buy half a package or something like that).

Then we conclude that the correct option is the first option.

If you want to learn more about domains, you can read:

brainly.com/question/1770447

#SPJ1

6 0

2 years ago

Can you help thanks this is my last one!​

Can you help thanks this is my last one!