Gcf of 63 and 27 i will give toy 12pts

You are given the dollar value of a product in 2015 and the rate at which the value of the product is expected to change during

Charra [1.4K]

Answer:

V = $3.50t + $90.5....

Step-by-step explanation:

V(t) is a function of t that expresses the value in year 2000+t.

We know that the increase is $3.50 times t.

So,

V(t) = $3.50t + c

where c is the constant.

V(15) = $3.50 (15) + c = $143 [t=15 as mentioned in the question]

and therefore

c = $143 - $3.50 (15)

c= $143 - $52.50

c= $90.5

Now we got the value of c. We can write the equation as

V = $3.50t + $90.5....

4 0

3 years ago

Todd is painting a banner on a wall in the cafeteria. The banner is in the shape of a rectangle and has a blue background. Todd

anastassius [24]

Well todd cant use all of the paint because that would be just waisting it because he has enough paint to cover the banner

6 0

3 years ago

Evaluate 4(a^2 + 2b) - 2b when a = 2 and b = –2.

Rama09 [41]

Answer:

4

Step-by-step explanation:

4(4-4)+4=4

insert 2 in a and-2 in b

7 0

3 years ago

Read 2 more answers

Write an equation of a parabola that passes through (3,-30) and has x-intercepts of -2 and 18. Then find the average rate of cha

Nookie1986 [14]

Answer:

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$ . The average rate of change of the parabola is -4.

Step-by-step explanation:

We must remember that a parabola is represented by a quadratic function, which can be formed by knowing three different points. A quadratic function is standard form is represented by:

$y = a\cdot x^{2}+b\cdot x + c$

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$a$ , $b$ , $c$ - Coefficients, dimensionless.

If we know that $(3, -30)$ , $(-2, 0)$ and $(18, 0)$ are part of the parabola, the following linear system of equations is formed:

$9\cdot a +3\cdot b + c = -30$

$4\cdot a -2\cdot b +c = 0$

$324\cdot a +18\cdot b + c = 0$

This system can be solved both by algebraic means (substitution, elimination, equalization, determinant) and by numerical methods. The solution of the linear system is:

$a = \frac{2}{5}$ , $b = -\frac{32}{5}$ , $c = -\frac{72}{5}$ .