What is 37 on base eight

Barry owns a music store, He recorded the sales of a newly launched album as part of his biweekly inventory. The table shows the

dimulka [17.4K]

Hi! Basically, the problem is asking us to find the values of a, b, and c in the equation $f(x)=a x^{2} +bx+c$ . Since we have three unknowns, we just need three equations. We can find these equations by using the data in the table.

First let's plug x = 0 and f(x) = 0.
$0=0+0+c$
$c=0$

Now that we know c, it's time to pick two more pairs. Let's plug-in (2,78) and (4,152)

$78=4a+2b$
$152=16a+4b$

Before proceeding with the process of eliminating one variable, let us first reduce both equations to their lowest terms. We divide the first equation by 2 and we divide the second one by 4.

$39=2a+b$
$38=4a+b$

Next, we subtract equation 2 from equation 1.

$1=-2a$
$a=-0.5$

Finally, we substitute the value of a to equation 2 to get the value of b.

$38=4(-0.5)+b$
$b=40$

Therefore, the function should be $f(x)=-0.5 x^{2} +40x$

5 0

3 years ago

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Simplify the expression 5+4+2(-3)+7+(-5)

liberstina [14]

Just use pemdas
(Parenthesis, exponents, multiplication, division, addition, subtraction)

4 0

3 years ago

Mkz-qetu-kwp j o i n a l l a l o u d

astraxan [27]

is this a meet code?

i would lime to join if so

6 0

3 years ago

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Kyle works at a donut factory, where a 10-oz cup of coffee costs 95¢, a 14-oz cup costs $1.15, and a 20-oz cup costs $1.5

Fynjy0 [20]

Answer:

Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.

Step-by-step explanation:

Let 10-oz, 14-oz, and 20-oz coffees be represented by the variables a, b, and c, respectively.

Since a total of 14 cups of coffee was served:

$a+b+c=14$

A total of 204 ounces of coffee was served. Therefore:

$10a+14b+20c=204$

A total of $16.70 was collected. Hence:

$0.95a+1.15b+1.5c=16.7$

This yields a triple system of equations. In order to solve a triple system, we should isolate the system to only two variables first.

From the first equation, let's subtract a and b from both sides:

$c=14-a-b$

Substitute this into both the second and third equations:

$10a+14b+20(14-a-b)=204$

And:

$0.95a+1.15b+1.5(14-a-b)=16.7$

In this way, we've successfully created a system of two equations, which can be more easily solved. Distribute:

For the Second Equation:

$\displaystyle \begin{aligned} 10a+14b+280-20a-20b&=204\\ -10a-6b&=-76\\5a+3b&=38\end{aligned}$

And for the Third:

$\displaystyle \begin{aligned} 0.95a+1.15b+21-1.5a-1.5b&=16.7\\ -0.55a-0.35b&=-4.3\end{aligned}$

We can solve this using substitution. From the second equation, isolate a:

$\displaystyle a=\frac{1}{5}(38-3b)=7.6-0.6b$

Substitute into the third:

$-0.55(7.6-0.6b)-0.35b=-4.3$

Distribute and simplify:

$-4.18+0.33b-0.35b=-4.3$

Therefore:

$-0.02b=-0.12\Rightarrow b=6$

Using the equation for a:

$a=7.6-0.6(6)=4$

And using the equation for c:

$c=14-(4)-(6)=14-10=4$

Therefore, Kyle filled 4 10-oz cups, 6 14-oz cups, and 4 20-oz cups.

7 0

2 years ago

Does the relation y=x^2-1 defines y as a function of x

Reil [10]

Answer:

Yes.

From Definition - A special relationship where each input has a single output.

Check if there is going to be one y for one x.

Step-by-step explanation:

Re-write it in terms of y=…., meaning solve for y
If it looks like you have two equations then it is a relationship -case a
If you need to prove that one has a function for case b, you would point out that if you substitute any value you will get some value in Reals.

7 0

3 years ago

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