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1 year ago

54 equals 9 more than a number t .

Mathematics

1 answer:

aleksandr82 [10.1K]1 year ago

6 0

Answer:

t=45

Step-by-step explanation:

54=t+9

54-9=t

t=45

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1 year ago

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Emily and a friend bought two tickets to see a soccer game. Each ticket cost $8.25. The friends paid a total of $24.40, which in

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$7.90

Step-by-step explanation:

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2 years ago

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

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2 years ago

A boat tour guide expects his tour to travel at a rate of x mph on the first leg of the trip. On the return route, the boat trav

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The intervals included in solution are:

$\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0$

Solution:

Given that,

A boat tour guide expects his tour to travel at a rate of x mph on the first leg of the trip

On the return route, the boat travels against the current, decreasing the boat's rate by 10 mph

The group needs to travel an average of at least 24 mph

Given inequality is:

$\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}$

We have to solve the inequality

$\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}\\\\\frac{2}{x} - 10\geq \frac{2}{24}$

$\mathrm{Subtract\:}\frac{2}{24}\mathrm{\:from\:both\:sides}\\\\\frac{2}{x}-10-\frac{2}{24}\ge \frac{2}{24}-\frac{2}{24}\\\\Simplify\\\\\frac{2}{x}-10-\frac{2}{24}\ge \:0$

$\frac{2}{x}-\frac{10}{1}-\frac{2}{24} \geq 0\\\\\frac{ 2 \times 24}{x \times 24} -\frac{10 \times 24}{1 \times 24} - \frac{2 \times x }{24 \times x}\geq 0\\\\\frac{48}{24x}-\frac{240x}{24x}-\frac{2x}{24x}\geq 0\\\\Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions\\\\\frac{48-240x-2x}{24x}\geq 0\\\\Add\:similar\:elements\\\\\frac{48-242x}{24x}\ge \:0$

$\mathrm{Multiply\:both\:sides\:by\:}24\\\\\frac{24\left(48-242x\right)}{24x}\ge \:0\cdot \:24\\\\Simplify\\\\\frac{48-242x}{x}\ge \:0\\\\Factor\ common\ terms\\\\\frac{-2\left(121x-24\right)}{x}\ge \:0\\\\\mathrm{Multiply\:both\:sides\:by\:}-1\mathrm{\:\left(reverse\:the\:inequality\right)}$

When we multiply or divide both sides by negative number, then we must flip the inequality sign

$\frac{\left(-2\left(121x-24\right)\right)\left(-1\right)}{x}\le \:0\cdot \left(-1\right)\\\\\frac{2\left(121x-24\right)}{x}\le \:0\\\\\mathrm{Divide\:both\:sides\:by\:}2\\\\\frac{\frac{2\left(121x-24\right)}{x}}{2}\le \frac{0}{2}\\\\Simplify\\\\\frac{121x-24}{x}\le \:0$

$\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{121x-24}{x}\\$

This is attached as figure below

From the attached table,

$\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\le \:\:0\\\\0$

Therefore, solution set is given as:

$\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0$

8 0

2 years ago

Read 2 more answers