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3 years ago

Choose the graph which represents the solution to the inequality: -2x + 5 ≤ 11

Mathematics

1 answer:

masha68 [24]3 years ago

4 0

Answer:

Step-by-step explanation:

-2x + 5 </= 11

-2x </= 6

x =/> -3

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Which problem situation could the equation below represent? 164/x = 8 (x ≠0)

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Write each expression as a single natural logarithm. ln 3+ ln 4

Oduvanchick [21]

Expression as a single natural logarithm is 2.48490664979.

<h3>What is natural logarithm?</h3>

ln 3+ ln 4

ln (3×4)

ln(12) =2.48490664979

The inverse of an exponential function, the natural log is the logarithm to the base of the number e. Natural logarithms are unique varieties of logarithms that are employed in the treatment of time and growth-related issues.

The distinction between log and ln is that log is expressed in terms of base 10, while ln is expressed in terms of base e. As an illustration, log of base 2 is denoted by log2 and log of base e by loge = ln (natural log).

To learn more about natural logarithm from the given link:

brainly.com/question/20785664

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1 year ago

Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$