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ANTONII [103]
3 years ago
10

Jake has $11 left in his pocket. he spent $3 on a bus ticket, $5 on lunch, and $5 on a movie. how much money did he have at the

beginning of the day?
Mathematics
2 answers:
KATRIN_1 [288]3 years ago
4 0
$24 if it is at the beginning of the day and he is left with $11, and if he spent $13, then 11+13=24
iragen [17]3 years ago
4 0

Answer:

Jake had twenty-four dollars ($24) at the beginning of the day.

Step-by-step explanation:

Jake has $11 dollars left in his pocket, after spending 3 dollars on a bus ticket, 5 dollars on lunch, and 5 more dollars on a movie. 3+5+5 is equal to 13, therefore if after spending 13 dollars Jake still has 11 dollars, which means that at the beginning of the day Jake had 24 dollars (11+13).

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Simply the following expression 2^x - 2^x + 1​
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Answer:

1

Step-by-step explanation:

the 2^x - 2^x cancels out

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3 years ago
(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla
natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}

    =\sqrt{1+1}\\\\=\sqrt{2}

calculating unit vector:

\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}

the point Q is at a distance h from P(6,6) Here, h=0.1  

a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071

the value of Q= (5.92928 ,6.07071  )

In point b:

Calculating the directional derivative of f (x, y) = \sqrt{x+3y} at P in the direction of \vec{v}

f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\

            =\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557

\vec{v} = 1.97651557

In point C:

Computing the directional derivative using the partial derivatives of f.

f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot   \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}

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A puppy and kitten are 180 meters apart when they see each other. The puppy can run at a speed of 25 meters per second, while th
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It seems it should be the other way... . How soon will the puppy catch up with the kitten.

The way you have it the kitten will never catch the pup. There is already 180m between them so, the distance will only increase because the puppy runs faster than the cat.

If the puppy runs after the cat, will eventually catch up with it and can be calculated. Not the other way around.

If the pup runs after the cat, the distance between them will decrease and the pup will catch the cat.

RT = D kitten

RT = D+180 puppy

20T = D

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It equals 96. hope that helped
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