What is -3/4(2/3 in simplest form

inn [45]

X^2+10x+16＝0

First, you find factors of 16.
1 x 16
2 x 8
4 x 4

Next, you find which of the factor pairs adds up to 10 (from the 10x). In this case, 2 x 8 because 2 times 8 is 16 and 2 plus 8 is 10.

Then, the equation will be written out as: (x+2) (x+8). Take those two equations and set them equal to 0, and then solve.

x+2＝0
-2 -2
x＝-2

x+8＝0
-8 -8
x＝-8

So, your answers are x＝-2 and x＝-8. You can check by plugging in those two numbers as x.

7 0

3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

An item travels 70 ft in 10 s. What is the unit rate?

sveticcg [70]

Distance = (rate)(time)

if you want o find the rate you have to get in by itself on one side so divide both sides by time

distance/time = rate

now replace the words with the values and complete the problem

(70 ft)/(10s) = rate
7 feet per sec = rate

The answer is that the unit rate is 7 feet per sec (or fps)

5 0

3 years ago

Please help its my last question and I really need to submit this right about now. I’m not sure what the answer is and how I do

iren [92.7K]

Answer:

I legit answered this its 6/10

Edit: y = -6/10x + 100

Step-by-step explanation:

Change in y over change in x

60/100

6/10

8 0

3 years ago

Read 2 more answers

How do you use compatible numbers to estimate the quotient $595 divided by 25

storchak [24]

Well, 25 certainly doesn't divide evenly into 595, but it does divide into 600, which is <em>very</em> close to 595. So if we divide 600 by 25, we can get a rough estimate of what 595 divided by 25 is.
$\frac{600}{25} = 6* \frac{100}{25} = 6* 4=24$

4 0

3 years ago