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nignag [31]
2 years ago
9

A cab company charges $2 per ride when the passengers get into the vehicle, then each mile is an additional $1.50. Write an equa

tion to represent this scenario.
Mathematics
2 answers:
SpyIntel [72]2 years ago
4 0

Answer:

2 + 1.5x

Step-by-step explanation:

$2 to start, then $1.50 for x amount of miles, so you would just substitute whatever number of miles instead of x to get the total charge

Alex787 [66]2 years ago
3 0
This scenario is an example of a linear equation. Therefore, we can use slope-intercept form to represent the equation.

Slope-Intercept Form: y=mx+b
where m=slope and b=y-intercept.

In this scenario, m=1.50, because the rate is $1.50/mile, so the line will increase $1.50 as each mile is added. The y-intercept or b=2 because that is the constant starting point. Passengers are immediately charged 2 dollars for a ride, which never changes, it’s just a 2 dollar charge. Therefore, b=2.

Putting the equation into y=mx+b

y=1.50x+2 is your answer. The line will start at $2 and rise from there $1.50 per each mile.

Answer: y=1.50x+2

Please mark Brainliest!!!
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Describe the steps required to determine the equation of a quadratic function given its zeros and a point.
faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

y = a\cdot x^{2}+b\cdot x + c (1)

Where:

x - Independent variable.

y - Dependent variable.

a, b, c - Coefficients.

A value of x is a zero of the quadratic function if and only if y = 0. By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: A(x,y) = (r_{1}, 0), B(x,y) = (r_{2},0) and C(x,y) = (x,y)

Based on such information, we form the following system of linear equations:

a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0 (2)

a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0 (3)

a\cdot x^{2} + b\cdot x + c = y (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }

c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}

And finally we obtain the equation of the quadratic function given two zeroes and a point.

6 0
3 years ago
6. Fill in the smallest digit to make the number divisible by:
Degger [83]

Answer:

(i) 7164<u>0</u>, 32197<u>0</u>

(ii) 1<u>1</u>43, 47<u>2</u>05, <u>2</u>316

(iii) <u>1</u>428, 9<u>2</u>52, 721<u>2</u>

(iv) 2462<u>0</u>, 91<u>00</u>, 670<u>0</u>

(v) 1232<u>0</u>, 59<u>0</u>16, 4642<u>4</u>

Step-by-step explanation:

(i) In order for any number to be divisible by 5, its last digit must be <u>either 0 or 5</u>.

In this case, we should add a last digit, so we opt for <u>0</u> without any doubt since it gives a smaller number than <u>5</u> as the last digit: 71640 is smaller than 71645, the same way as 321970 is smaller than our second option, 321975.

(ii) If the sum of a number's digits gives a number divisible by 3, then the main number is also divisible by 3.

This means that we have to do some addition:

1_43: 1 + 4 + 3 = 8. Eight is not divisible by 3. But <u>9</u> is. So the smallest digit we can add for it to be divisible by three is <u>1</u>: <em>1143.</em>

47_05: 4 + 7 + 0 + 5 = 16. Since 16 is not, the next number that is divisible by 3 is 18, and therefore, the smallest digit we need here is <u>2</u>: <em>47205.</em>

_316: 3 + 1 + 6 = 10. Twelve is the next number divisible by 3, and that means <u>2</u> is the digit we choose: <em>2316.</em>

(iii) Only numbers that are divisible by both <u>2 and 3</u> are also divisible by 6.

_428: In order for a number to be divisible by 2, its last digit must be 0 or divisible by 2. That is already the case here.

4 + 2 + 8 = 14 which is not divisible by 3. But 15 is. So the smallest digit we need is <u>1</u>: <em>1428.</em>

9_52: The number will be divisible by 2. And 9 + 5 + 2 = 16 which is not divisible by 3 but 18 is. This means the smallest digit we need is <u>2</u>: <em>9252.</em>

721_: Since the last digit can be any of these <u>0, 2, 4, 6, 8</u> for the number to be divisible by two, we will first see what digit we need for it to be divisible by 3.

7 + 2 + 1 = 10 which is not divisible by 3 but 12 is. We will then choose to add <u>2</u>: <em>7212.</em>

(iv) A number is divisible by 4 only if the number that is formed by its two last digits is also divisible by 4..

2462_: Since we need to add a last digit, this should be easy. The smallest two-digit-number that starts with 2 and is divisible by four is <u>20</u>: <em>24620. </em>

91__: Here we need to add both of the two last digits, and this should be the easiest example! All numbers that end in <u>00</u>, meaning they have a round number of hundreds, are divisible by 4! Therefore our option is: <em>9100.</em>

670_: In this case, we apply the same rule from the last example - and we choose <u>0</u>: <em>6700.</em>

(v) In order for a number to be divisible by 8, its last three digits need to be divisible by 8, or it can end in <u>000,</u> since every number having a round number of thousands is also divisible by 8.

1232_: 320 is divisible by 8 since 8 × 40 gives 320. Therefore, our smallest digit here is <u>0</u>: <em>12320.</em>

59_16: If we add a <u>0</u> here, the last three digits form the number 016, which is divisible by 8. Our choice is again zero: <em>59016.</em>

4642_: Unfortunately, 420 is not divisible by 8. The next number closest to 420 and which is divisible by 8 is 424. Its division by 8 gives 53. Therefore we should add <u>4</u>: <em>46424.</em>

7 0
3 years ago
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