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3 years ago

Solve for the variable in the following inequality 18 ≥ 6x

Mathematics

1 answer:

baherus [9]3 years ago

6 0

Answer:

3 > x

Step-by-step explanation:

(Sorry if im wrong)

Divide both sides by 6

18/6 > 6x/6

3 > x

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A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1

Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth $= \$ 17,000$

depreciates by $\frac{1}{2}$ in 3 years.

time= 19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$\to A_t=A_0(1-\frac{r}{100})^t$

calculates depreciate value in 3 year $= \frac{1}{2} \times 17,000$

$= 8,500$

so,

$A_t=8,500\\\\A_0=17,000\\\\t=3\ years$

$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238$

4 0

3 years ago

Read 2 more answers

Carly has 15 stuffed animals. How many ways are there to organize her top 3 favorites?

Ivanshal [37]

Well, the answer will depend on whether the order will count or not (based on Permutations and Combinations). If the order counts, then we would use the formula for Permutations, which is:
$\frac{n!}{(n-r)!}$
Where n is the number of items you have, and r is the number of times you choose from the items.
$\frac{15!}{(15-3)!}$
Which simplifies to
$\frac{15!}{(12)!}$
Which simplifies to 15*14*13 (because all the numbers 1-12 in the factorial canceled out), which gets us the answer 2730.

Now, if you wanted to find the number of ways to order the toys without replacement (order doesn't count), you would use the formula:
$\frac{n!}{r!(n-r)!}$
The variables are still the same, but you are now multiplying by r!.
$\frac{15!}{3!(15-3)!}$
Simplifies to
$\frac{15!}{3!(12)!}$
Which simplifies to (using the same cancellation method above)
$\frac{2730}{3!}$
Dividing 2730 by 3! will get us an answer of 455.

Really, it depends on whether they are ordered or not. In this case (since you didn't specify whether the order mattered), the answer would be 455 or 2730.

:)

4 0

3 years ago

How do i check an equation for extraneous solutions?

Marina86 [1]

Example:

$\sqrt x(x+1)=0$

This suggests two solutions, $x=0$ and $x=-1$ .

However, upon plugging these solutions back into the equation, you get

$\sqrt0(0+1)=0(1)=0$

which checks out, but

$\sqrt{-1}(1-1)=0$

does not because $\sqrt x$ is defined only for $x\ge0$ (assuming you're looking for real solutions only). So, we call $x=-1$ an extraneous solution, and the complete solution set (over the real numbers) is $x=0$ .

7 0

3 years ago

Read 2 more answers

During a weekend, the manager of a mall gave away gift cards to every 80th person who visited the mall. On Saturday, 1,210 peopl

Talja [164]

Answer:

Step-by-step explanation:

7 0

3 years ago

I need help with this question

Temka [501]

Answer: