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galina1969 [7]
2 years ago
14

Gas mileage is the number of miles you can drive on a a gallon of gasoline. A test of a new car results in 510 miles on 10 gallo

ns of gas. How far could you drive on 40 gallons of gas? What is the car's gas mileage
a. 2040 miles and 51 miles/gallon
b. 2050 miles and 200 miles/ gallon
c. 3040 miles and 50 miles/ gallon
d. 230 miles and 51 miles/gallon
help plis
Mathematics
1 answer:
hammer [34]2 years ago
3 0

Answer:

a. 2040 miles and 51/gallon

Step-by-step explanation:

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Find the solutions to 2x⁴-24x²+40=0 and the x-intercepts of the graph of y=2x⁴-24x²+40.
jekas [21]

Answer:

The solutions and the x-intercepts of the polynomial 2x^4-24x^2+40 are:

x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}

Step-by-step explanation:

Given a function <em>f</em> a solution or a root of <em>f</em> is a value x_{0} at which f(x_0)=0.

An x-intercept is a point on the graph where y is zero.

To find the solutions of the polynomial and the x-intercepts 2x^4-24x^2+40 you need to:

First, we need to factor the polynomial expression

Factor the common term

{\left(2 x^{4} - 24 x^{2} + 40\right)} = {\left(2 \left(x^{4} - 12 x^{2} + 20\right)\right)}

We can treat x^{4} - 12 x^{2} + 20 as a quadratic function with respect to x^2

Let u=x^2. We can rewrite x^{4} - 12 x^{2} + 20 in terms of u as follows:

u^2-12u+20

We need to solve the quadratic equation

u^2-12u+20=0

for this we can use the Quadratic Equation Formula:

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-12,\:c=20:\quad u_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}

u_1=\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_1=10

u_2=\frac{-\left(-12\right)-\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_2=2

the solutions to the quadratic equation are:

u=10,\:u=2

Therefore, u^2-12u+20=(u-10)(u-2)

Recall that u=x^2 so

2 x^{4} - 24 x^{2} + 40=2 \left(x^{2} - 10\right) \left(x^{2} - 2\right)=0

Using the Zero factor Theorem: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

x^2-10=0 roots are x_1=\sqrt{10}; x_2=-\sqrt{10}

x^{2} - 2=0 roots are x_1=\sqrt{2}; x_2=-\sqrt{2}

The solutions and the x-intercepts are:

x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}

Because all roots are real roots the x-intercepts and the solutions are equal.

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