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vlabodo [156]
3 years ago
11

Mike recently increased the size of his Jeep tires from the original 29 inch diameter to the larger 33.73 inch diameter. If Mike

didn't recalibrate his speedometer, how fast is he really going on the new tires when his speedometer shows he is traveling 60 mph?
a. 54.5 mph
b. 62.1 mph
c. 66.1 mph
d. 69.8 mph
Mathematics
2 answers:
Anastasy [175]3 years ago
4 0

Answer:

d. 69.8 mph

Step-by-step explanation:

Since, the ratio of the diameter of the tyre of a vehicle and its speed must be constant,

Given,

The original diameter of the tyre = 29 inch,

Original speed = 60 mph,

Thus, the ratio of diameter and the speed of the vehicle = \frac{29}{60}

New diameter of the tyre = 33.73 inch,

Let x be the new speed of the vehicle = \frac{33.73}{x}

\implies \frac{29}{60}=\frac{33.73}{x}

\implies x=\frac{33.73\times 60}{29}=69.79\approx 69.8\text{ mph}

Hence, the actual speed of the vehicle would be 69.8 mph.

OPTION D is correct.

coldgirl [10]3 years ago
3 0

Answer:

I believe it is C

Step-by-step explanation:

Please correct me if i am wrong ^w^

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Answer:

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Step-by-step explanation:

2.5 h

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12x2.5=30

6 0
3 years ago
225,120,000 in scientific notation
KIM [24]
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5 0
4 years ago
if a store uses a selling price based markup of 40% and an item cost the store $300 what selling price would the store set for t
kirza4 [7]

Answer:

The selling price would be $420.

Step-by-step explanation:

In order to find the mark up, we need to multiply the amount it originally costs by the percentage it is being marked up.

$300 * 40% = $120

Now that we have the mark up amount, we add it to the original cost to get the sale price.

$300 + $120 = $420

6 0
3 years ago
If alpha and beta are the roots of a quadratic polynomial 3x^2-6x-1 find the values of (Alpha-beta)
Murrr4er [49]

Answer:

Either -4\sqrt{6} or 4\sqrt{6}, depending on whether \alpha is larger than \beta.

Step-by-step explanation:

The two roots (might necessarily be distinct or real) of the quadratic equation

ax^{2} + bx + c = 0, where a, b, and c are constants and a\ne 0 are

  • \displaystyle x_1 = \frac{-b+\sqrt{\text{b^{2} - 4ac}}}{2a}, and
  • \displaystyle x_2 = \frac{-b-\sqrt{\text{b^{2} - 4ac}}}{2a}.

The difference between the two will be either:

x_1 - x_2 = 2\sqrt{b^{2} - 4ac} or

x_2 - x_1 = -2\sqrt{b^{2} - 4ac}.

For this question,

  • a = 3,
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x_1 - x_2 = 2\sqrt{(-6)^{2} - 4\times 3\times (-1)} = 4\sqrt{6}, or

x_1 - x_2 = -2\sqrt{(-6)^{2} - 4\times 3\times (-1)} = -4\sqrt{6}.

7 0
3 years ago
Hello, can someone help me with this? Thank you.
Ksivusya [100]

Answer:

(x, y) → (2x - 2, 2y - 1)

Step-by-step explanation:

The original length from Point E to Point G was 2. The length after the figure was transformed is 4. This means that the value of x was multiplied by 2 because 2(the original value) multiplied by 2 equals 4(the new value.)

After the x value is doubled, Point E is found at x = 0 when the transformed Point E is at x = -2. This tells us that the x value has to decrease by 2 after it is doubled.

The original length from Point F to Point H was 4. The length after the figure was transformed is 8. This means that the y value doubled in value, which is the same as saying it was multiplied by 2.

After you double the y value, Point F is found at y = 9, but the fully transformed figure's Point F is found at y = 8. This means that after the y value is doubled it should decrease by 1.

I hope this helps, and let me know if you have any other questions :)

3 0
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