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3 years ago

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Domain and range of g(x)= 5x-3/2x+1 Solve for domain and range?

1 answer:

alukav5142 [94]3 years ago

3 0

The domain and the range

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Abdul invests $5,000 at an interest rate of 4%, compounded quarterly. How much is the investment worth at the end of 3 years?

notsponge [240]

I think is b !!!!!!!

8 0

3 years ago

A box of chalk and 2 staplers cost $10. three boxes of chalk and 2 staplers cost $18. find the total cost of 1 box of chalk and

Alchen [17]

Here is the solution of the given problem above:
Let s= staples
<span>c = chalk</span>

<span>c + 2s = 10 </span>
<span>3c + 2s = 18 </span>

<span>-c-2s = -10 </span>
<span>3c + 2s = 18 </span>

<span>2c = 8 </span>

<span>c = 4 </span>
<span>s = 3
</span>Therefore, the price per chalk box is $4 and per stapler is $3.
So the total cost of the chalk and stapler would be $7.
Hope this is the answer that you are looking for. Thanks for posting it here in brainly.

7 0

4 years ago

Use trigonometric ratios to solve for X to the nearest tenth

Elan Coil [88]

Answer: x = 106.5

<u>Step-by-step explanation:</u>

You can use either sin or cos. I will use cos.

$cos\ \theta=\dfrac{adjacent}{hypotenuse}\\\\\\cos\ 62^o=\dfrac{50}{x}\\\\\\x\ cos\ 62^o=50\\\\\\x=\dfrac{50}{cos\ 62^o}\\\\\\.\quad=\dfrac{50}{0.469}\\\\\\.\quad=106.5$

6 0

4 years ago

Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year, write an exponential model for the

elena55 [62]

Answer:

In 2026 car will have a value of $5,000.

Step-by-step explanation:

We have been given that Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year.

Since we know that an exponential function is in form: $y=a*b^x$ , where,

a = Initial value,

b = For decay or decrease b is in form (1-r), where r represents decay rate in decimal form.

Let us convert our given decay rate in decimal form.

$14\%=\frac{14}{100}=0.14$

Upon substituting a =28,000 and r=0.14 in exponential decay function we will get,

$y=28,000(1-0.14)^x$ , where x represents number of years after 2014.

Therefore, the function $y=28,000(0.86)^x$ represents the value of car x years after 2014.

To find the number of years it will take to car have the value of $5,000, we will substitute y=5,000 in our function.

$5,000=28,000(0.86)^x$

Let us divide both sides of our equation by 28,000.

$\frac{5,000}{28,000}=\frac{28,000(0.86)^x}{28,000}$

$0.1785714285714286=(0.86)^x$

Let us take natural log of both sides of our equation.

$ln(0.1785714285714286)=ln((0.86)^x)$

Using natural log property $ln(a^b)=b*ln(a)$ we will get,

$ln(0.1785714285714286)=x*ln(0.86)$

$\frac{ln(0.1785714285714286)}{ln(0.86)}=\frac{x*ln(0.86)}{ln(0.86)}$

$\frac{-1.7227665977411033893}{-0.1508228897345836}=x$

$x=11.422447\approx 12$

As in the 12th year after 2014 car will have a value of $5,000, so we will add 12 to 2014 to find the year.

$2014+12=2026$

Therefore, in 2026 car will have a value of $5,000.

6 0

3 years ago

Read 2 more answers

?!!!!!!!!!!!!!!!!!!!!!!!

Phoenix [80]

The correct answer is B.

Can I please have brainliest?

8 0

3 years ago