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vladimir2022 [97]
3 years ago
8

A bus company sell an annual bus pass for $8.50 and then charges $0.25 per ride.Use the equation c=0.25b + 8.50 to find the coas

t for someone who rides the bus 38 times.
Help me
Mathematics
1 answer:
Dmitrij [34]3 years ago
4 0
The answer is $18. Put the 38 in the place of b, b being the amount of times someone rides the bus. Hence, C=0.25(38) + 8.50. Then, multiply the multiplable numbers. You get 9.5. Add 9.5 and 8.50 to get 18.
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In this exercise, we want to know the x-intercepts of each item. To find the x-intercepts, set y = 0 as indicated in each item and solve for x. So:

<h2>1. Answer:</h2>

B. x=-1; x=-1.75

<h3>Step by step explanation:</h3>

we have the equation:

4x^2+11x+7=0

We can say that this equation comes from the function f(x)=4x^2+11x+7 so we have set y=0 to find the x-intercepts. By using the quadratic formula we have:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ where: \\ \\ a=4, \ b=11, \ c=7 \\ \\ x=\frac{-11 \pm \sqrt{11^2-4(4)(7)}}{2(4)} \\ \\ x=\frac{-11 \pm \sqrt{121-112}}{8} \\ \\ \boxed{x_{1}=-1 \ and \ x_{2}=-1.75}

<h2>2. Answer:</h2>

B. x=-1; x=-1.75

<h3>Step by step explanation:</h3>

we have the equation:

3x^2-4x+1=0

We can establish a function g(x)=3x^2-4x+1 and say that we want to find the x-intercepts of this function by setting y = 0. Therefore, by using the quadratic formula we have:

a=3, \ b=-4, \ c=1 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{2(3)} \\ \\ x=\frac{4 \pm \sqrt{16-12}}{6} \\ \\ \boxed{x_{1}=1 \ and \ x_{2}=\frac{1}{3}}

<h2>3. Answer:</h2>

H. No Solution

<h3>Step by step explanation:</h3>

we have the equation:

3x^2-4x+2=0

We can establish a function h(x)=3x^2-4x+2 and say that we want to find the x-intercepts of this function by setting y = 0. Therefore, by using the quadratic formula we have:

a=3, \ b=-4, \ c=2 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(3)(2)}}{2(3)} \\ \\ x=\frac{4 \pm \sqrt{16-24}}{6}

Since 16 - 24 = -8, that is, a number less than zero which is within a square root, we say that the equation 3x^2-4x+2=0 has no any real solution.

<h2>4. Answer:</h2>

E. x=1

<h3>Step by step explanation:</h3>

we have the equation:

x^2-2x+1=0

We can establish a function c(x)=x^2-2x+1. By setting y = 0 we'll find the x-intercepts. Let's solve this problem using other method. You can find some binomial products having a special form. So it's easier to find a solution by using distributive. The form of this polynomial is a Square of a Binomial in the form:

(x-1)^2=0 \\ \\ Because: \\ \\ (x-1)^2=(x-1)(x-1)=x^2-x-x+1= x^2-2x+1

Therefore, the value that satisfies this equation is \boxed{x=1}

<h2>5. Answer:</h2>

K. x = -1

<h3>Step by step explanation:</h3>

we have the equation:

x^2+2x+1=0

We can establish a function a(x)=x^2+2x+1. By setting y = 0 we'll find the x-intercepts. We are going to solve this problem by using the previous method. The form of this Square of a Binomial is:

(x+1)^2=0 \\ \\ Because: \\ \\ (x+1)^2=(x+1)(x+1)=x^2+x+x+1= x^2+2x+1

Therefore, the value that satisfies this equation is \boxed{x=-1}

<h2>6. Answer:</h2>

N) x = 1/2

<h3>Step by step explanation:</h3>

we have the equation:

4x^2-4x+1=0

We can establish a function b(x)=4x^2-4x+1 and say that we want to find the x-intercepts of this function by setting y = 0. Here we will use the quadratic formula, so:

a=4, \ b=-4, \ c=1 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(4)(1)}}{2(4)} \\ \\ x=\frac{4 \pm \sqrt{16-16}}{8} \\ \\ \boxed{x=\frac{1}{2}}

So we have just one solution.

<h2>7. Answer:</h2>

M) x = -1/2

<h3>Step by step explanation:</h3>

we have the equation:

4x^2+4x+1=0

We can establish a function b(x)=4x^2+4x+1 and say that we want to find the x-intercepts of this function by setting y = 0. As in the previous exercise, we will use the quadratic formula, so:

a=4, \ b=4, \ c=1 \\ \\ x=\frac{-4 \pm \sqrt{(4)^2-4(4)(1)}}{2(4)} \\ \\ x=\frac{-4 \pm \sqrt{16-16}}{8} \\ \\ \boxed{x=-\frac{1}{2}}

So we have just one solution.

<h2>8. Answer:</h2>

D) x = -1.45; x=1.25

<h3>Step by step explanation:</h3>

we have the equation:

5x^2+x-9=0

We can establish a function D(x)=5x^2+x-9 and say that we want to find the x-intercepts of this function by setting y = 0. By using the quadratic formula we can solve this problem, so:

a=5, \ b=1, \ c=-9 \\ \\ x=\frac{-1 \pm \sqrt{(1)^2-4(5)(-9)}}{2(5)} \\ \\ x=\frac{-1 \pm \sqrt{1+180}}{10} \\ \\ \boxed{x_{1}=-1.45 \ and \ x_{2}=1.25}

<h2>9. Answer:</h2>

J) x = 4; x=-3

<h3>Step by step explanation:</h3>

we have the equation:

-x^2+x+12=0

We can establish a function k(x)=-(x^2-x-12) and say that we want to find the x-intercepts of this function by setting y = 0. In this exercise we'll use other method. Since this is a non-perfect square trinomial, we know that:

(x+a)(x+b)=x^2+(a+b)x+ab

So let's find two numbers such that the sum is -1 and the product is -12. Those numbers are -4 and 3, thus:

-(x-4)(x+3)=-x^2+x+12=0

Therefore, our solutions are:

x=4 \ and \ x=-3

________________

<h3>THE OTHER SOLUTIONS HAVE BEEN ATTACHED BELOW</h3>
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6 0
3 years ago
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