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1 year ago

D=vit + at solve for vi

Mathematics

1 answer:

gavmur [86]1 year ago

7 0

Answer:

$\large\boxed{vi=\dfrac{d-at}{t}}$

Step-by-step explanation:

$d=vit+at\qquad\text{subtract}\ at\ \text{from both sides}\\\\d-at=vit+at-at\\\\d-at=vit\qquad\text{divide obth sides by}\ t\\\\\dfrac{d-at}{t}=\dfrac{vit}{t}\\\\\dfrac{d-at}{t}=vi$

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Mark has a 25 ounce drink. He drinks 5 ounces. What percent of the drink does he have left?

valina [46]

Answer: $80\%$

Step-by-step explanation:

This problem can be solved by the Rule of Three, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known.

In this case, the 25 ounce drink represents the $100 \%$ . So, if Mark drinks 5 ounces, this means he has 20 ounces left and we have to find the percent this represents:

$25 oz$ ------- $100\%$

$20 oz$ -------- $?$

$?=\frac{(20 oz)(100\%)}{25 oz}$

$?=80\%$ This is the percent of the drink Mark has left

4 0

2 years ago

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0

2 years ago

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch costs $4.25. The total cost of the lunches is $

andriy [413]

Answer:

11 boxed lunches

Step-by-step explanation:

Full question

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch cost 4.25. The total cost of the lunches is 53.75, including a 7$ delivery fee. Write and solve an equation to find x the number of boxed lunches Janie ordered

First of all subtract the delivery feesince it was inckuded in the total cost, this will now be the total cost of all the noxed lunches ordered by Janie, then divide the balance of the total cost by the cost of one boxed lunch to get thd total boxed kunches

X= 53.75-7/4.25

X= 53.75-7= 46.75/4.25

X=11

7 0

1 year ago

Read 2 more answers

Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $

nlexa [21]

Answer:

$4.84

Step-by-step explanation:

Given that

For delivery within $1\frac{1}2$ mi, charges = $1.25

For delivery within $1\frac{1}2$ mi - $1\frac{3}4$ mi, charges = $1.70

For delivery within $1\frac{3}4$ mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2 $\frac{1}4$ mi, charges = $2.60

For delivery within 2 $\frac{1}4$ mi - 2 $\frac{1}2$ mi, charges = $3.05

For delivery within 2 $\frac{1}2$ mi - 2 $\frac{3}4$ mi, charges = $3.50

For delivery within 2 $\frac{3}4$ mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3 $\frac{1}4$ mi, charges = $4.40

So, every 0.25 mi or $\frac{1}4$ mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of $3\frac{1}8$ miles lies within the range 3 mi - 3 $\frac{1}4$ mi.

So actual charges after increase of 10%

$\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}$

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2 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit 10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is $\overline A=\textrm{Missing a target at least once}$

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, $P(A)=0.95^{10}=0.5987$

We know that the sum of probability of an event and its complement is 1.

So, $P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%$

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

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1 year ago

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