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motikmotik
2 years ago
5

The Allied Taxi Company charges $2.50 to pick up a passenger and then adds $1.95 per mile. Isaac was charged $27.46 to go from o

ne city to another. If x represents the number of miles driven by the taxi, which linear equation can be used to solve this problem, and how many miles did Isaac travel, rounded to the nearest tenth?
A.)1.95x + 2.50 = 27.46; Isaac traveled 15.4 miles.
B.)1.95x + 2.50 = 27.46; Isaac traveled 12.8 miles.
C.)2.50x + 1.95 = 27.46; Isaac traveled 11.8 miles.
D.)2.50x + 1.95 = 27.46; Isaac traveled 10.2 miles.
Mathematics
2 answers:
jarptica [38.1K]2 years ago
5 0

I think the answer is d

Step-by-step explanation:

hope this helps

Lorico [155]2 years ago
4 0

option d is the correct answer hope this helps u

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Twenty more than four times a number is equal to the difference between 139 and three times the number. Find the number.
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Answer:

x = 17

Step-by-step explanation:

First, let's turn this into an equation:

twenty more ( +20 ) than four times a number ( 4x ) is equal ( = ) to the difference ( - ) between 139 and 3 times the number ( 3x ).

4x + 20 = 139 - 3x

Now let's solve:

Subtract 20 from each side:

4x = 119 - 3x

Add 3x to each side:

7x = 119

Divide each side by 7:

x = 17

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2 years ago
Crickets can jump with a vertical velocity of up to 14 ft/s. Which equation models the height of such a jump, in feet, after t s
Y_Kistochka [10]

Answer:

h = 147 - 16t^2.

Step-by-step explanation:

Use the following equation of motion

h = ut + 1/2 gt^2    where  h = height, u = initial velocity , g = acceleration due to gravity and t = the time.

So here we have:

h = 14t + 1/2 * -32 * t^2

h = 14t - 16t^2

8 0
3 years ago
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52

Step-by-step explanation:

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Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke
Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that \sqrt{3}=\frac{m}{n} where  m and n are non negative integers, and the fraction \frac{m}{n} is irreducible, i.e., the numbers m and n have no common factors.

Now, squaring the equality at the beginning we get that

3=\frac{m^2}{n^2} (1)

which is equivalent to 3n^2=m^2. From this we can deduce that 3 divides the number m^2, and necessarily 3 must divide m. Thus, m=3p, where p is a non negative integer.

Substituting m=3p into (1), we get

3= \frac{9p^2}{n^2}

which is equivalent to

n^2=3p^2.

Thus, 3 divides n^2 and necessarily 3 must divide n. Hence, n=3q where q is a non negative integer.

Notice that

\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}.

The above equality means that the fraction \frac{m}{n} is reducible, what contradicts our initial assumption. So, \sqrt{3} is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, r\in\mathbb{Q}, which is equivalent to say that r=\frac{m}{n} where  m and n are non negative integers. Also, assume that k\in\mathbb{Z}. So, we want to prove that k\cdot r\in\mathbb{Z}. Recall that an integer k can be written as

k=\frac{k}{1}.

Then,

k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}.

Notice that the product mk is an integer. Thus, the fraction \frac{mk}{n} is a rational number. Therefore, k\cdot r\in\mathbb{Q}.

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Write q=x+p and let us suppose that q is a rational number. So, we get that

x=q-p.

But the subtraction or addition of two rational numbers is rational too. Then, the number x must be rational too, which is a clear contradiction with our hypothesis. Therefore, x+p is irrational.

7 0
4 years ago
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