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pashok25 [27]
2 years ago
12

Can someone help me with this ?

Mathematics
1 answer:
Virty [35]2 years ago
3 0
Mark drove for 5 hrs
pranav drove for 3 hrs

Distance = speed x time
Pranav =50kmh x 3hrs
= 150 km
* note that they drove the same distance

Speed of mark=distance / time
=150 km / 5hrs
= 30 kmh
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Thirty-six and fifty-two thousandths as a decimal number
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In decimal form that would be 36.052
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3 years ago
Alberta Doan worked 6 hours at time-and-a-half pay and 3 ¼ hours at double-time pay. Her regular pay rate was $9.72 an hour. Wha
natka813 [3]

Answer:

Alberta's total overtime pay of the week was $ 145.80.

Step-by-step explanation:

Given that Alberta worked 6 hours at time-and-a-half pay, and 3 hours at double-time pay, and that the value of her regular work hour is $ 9.72, to determine the value of the pay of his overtime, the following equation must be performed:

6x1.5x9.72 + 3x2x9.72 = X

9x9.72 + 6x9.72 = X

87.48 + 58.32 = X

145.8 = X

Therefore, Alberta's total overtime pay of the week was $ 145.80.

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=
BARSIC [14]
<h3>Answer:  -2</h3>

======================================================

Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0
2 years ago
The picture below shows the graph of which inequality?
AleksAgata [21]
The answer is B. The dotted line tells you that the equations will be "greater than" or "less than". A solid line would indicate that it would be "greater than or equal to" or "less than or equal to". Then, since C and D can be eliminated, you look at which equations graphs the line. The answer must be B because the y-intercept is -1 (remember that the equation of a line is y = mx + b, where b is the y-intercept).
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3 years ago
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