Ask question

Ask question

All categories

3 years ago

9

How do you solve this equation

1 answer:

lesantik [10]3 years ago

6 0

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ e^{x-6}=\left( \cfrac{1}{e^4} \right)^{x+2}\implies e^{x-6}=\left( e^{-4} \right)^{x+2}\implies \stackrel{\textit{same bases, same exponents}}{e^{x-6}=e^{-4x-8}} \\\\\\ x-6=-4x-8\implies 5x=-2\implies x=-\cfrac{2}{5}$

You might be interested in

Rafael Drive 405 miles using 20 gallons of gas. At that rate, how many gallons of gas would he need to drive 243 miles?

ANEK [815]

This is how I'd set it up:
405/20 = 243/x
this means the mpg (miles / gallons) for 1 situation is the same as the mpg for the other situation (because the rate is the same)
then just cross multiply and solve:
405x=243*20 or 405x=4860 or x = 12
he would need 12 gallons of gas for 243 miles

3 0

3 years ago

Find the minimum or maximum of the function. Describe the domain/range in interval notation y=6x^2-1

Margarita [4]

max is at vertex

in form $y=ax^2+bx+c$

the x value of the vertex is $\frac{-b}{2a}$

given, $y=6x^2-1$

a=6, b=0

the x value of the vertex is -0/(2*6)=0

the y value is $y=6(0)^2-1=0-1=-1$

so vertex is at (0,-1)

since the value of a is positive, the parabola opens up and the vertex is a minimum value of the function

therefore that value is the smallest value the function can be

domain=numbesr you can use for x

range=numbesr you get out of inputting the domain

domain=all real numbers

range is {y | y≥-1} since y=-1 is the minimum

7 0

3 years ago

Which figure has the same area as the parallelogram shown below? A. A triangle with a base of 4 mm and a height of 20 mm B. A tr

Arisa [49]

<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of $80mm^{2}$ , according to the following formula, which works for all rectangles and parallelograms:

$A_{parallelogram}=(b)(h)$ (1)

Where $b$ is the base and $h$ is the height

The<u> area of a triangle</u> is given by the following formula:

$A_{triangle}=\frac{1}{2}(b)(h)$ (2)

So, for option A:

$A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}$

Now, the <u>area of a trapezoid </u>is:

$A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)$ (3)

For option B:

$A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}$

For option C:

$A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}$ >>>>This is the correct option!

For option D:

$A_{rectangle}=(30mm)(8mm)=240mm^{2} \neq 80mm^{2}$

<h2>Therefore the correct option is C</h2>

5 0

3 years ago

Write 77/20 as a terminating decimal

ahrayia [7]

77/20 = 3.85 <== terminating decimal

8 0

3 years ago

WHAT WOULD THIS BEEE

Neporo4naja [7]

No, the angles are roughly 34, 44, and 102

5 0

3 years ago