Graph the function. y = -6x2

You are trying to find how many students will fit in each bus or van, so the two variables used will be "b" to represent how many students can fit in a bus and "v" to represent how many students can fit in a van.

High school A used 1 van and 6 buses, so there will be 1"v" and 6"b" for 372 students.

High school B used 4 vans and 12 buses, so there will be 4"v" and 12"b" for 780 students.

Now, represent these in equations:

$v+6b = 372\\4v + 12b = 780$

We can use substitution to solve this system:

$v + 6b = 372$ can be rewritten as $v = 372 - 6b$ after subtracting 6b from both sides. You can then substitute this new value of "v" into the other equation to solve for "b":

$4(372-6b) + 12b = 780\\1488 - 24b + 12b = 780\\1488 - 12b = 780\\-12b = -708\\b = 59$

After solving for b, you can then substitute the new value of b into the other equation to find the value of v:

$v + 6(59) = 372\\v + 354 = 372\\v = 18$

After solving for both variables, you find that each bus can hold 59 students and each van can hold 18 students.

5 0

3 years ago

Read 2 more answers

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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3 0

2 years ago

Graph the function. y = -6x2 - 12x -1